Lecture 2: General Theory

Key Takeaways

so today in case you were thinking that there wasn't enough theorems on Tuesday we're going to approve a lot of them today on the general metric space Theory I I like this notion of a theory of the spaces this is a notation that um Dr Casey Rodriguez taught me Loosely speaking where you know you're just trying to encapsulate what are all the things we can say about the most fundamental objects in the space that you're considering and a number of them will essentially a number of the theorems that will prove are essentially the same as what we would picture in ukian space but we have to be slightly more careful sometimes and all of this theory is going to be building up to understanding something that you probably haven't seen before if you haven't done 100b known as compact sets so we're going to be building up to talking about compact sets but first we have to understand what are the fundamental Tools in our tool belt with metrics based Theory and I do want to Note One Thing even though last time I said we can put a metric on every set which is true not every set has a nice metric on it uh in case have any of youall Taken topology before no worries if not it's a class that usually comes after real analysis but in topology talk not every topological space is metrizable I.E there's not necessarily a metric on it that will make the topology nice and we'll talk about what topology is a little bit today so in case you're interested in what that statement is we'll come to it but the basic idea is even though we can put a metric on every set it's sometimes the trivial one is not that interesting so all right let's go ahead and jump into the metric Bas theory for today so first off we're going to be talking about convergent sequences just like we did in RN so throughout this let xn and YN be the sequences we're considering and the first statement theorem suppose xn converges to x what should we expect about the point that it converges to well in ukian space we knew that this limit point had to be unique and in fact we can prove that this must be the case uh so then X is unique to prove a statement like this we choose a different point Y in our set to consider so what we're going to show now is that convergent sequences have unique limit points so prove Suppose there exists a y in your metric base X oh I should say sequences in your metric base XD so metric Bas x with metric D and suppose that there exists a y and x such that X then converges to Y our end goal is to show then that X must be equal to Y in fact now to do this what do we know about equality for points well by the axioms of what we're defining a metric space to be two points are going to be the same if the distance is precisely zero it's an if and only a statement so that's what we want to prove we want to prove that the distance between X and Y is zero so how can we do that well firstly let's write down what the definitions of them converging to X and Y is respectively so xn converges to X means that for all Epsilon bigger than zero there exists an n in the natural numbers such that for all n bigger than or equal to n and I'll this N1 we have that the distance from xn to X is less than Epsilon this is the definition of convergent sequence that we're dealing with and similarly xn converges to Y if there exists an N2 in the natural numbers for this Epsilon such that for such that for all n bigger than or equal to N2 the distance from xn to Y is is less than Epsilon so now what do we want to do we want to use this to show that the distance between X and Y is small and to do this we'll apply the triangle inequality to understand the distance from X to Y so by the triangle inequality this is less than or equal to the distance from xn to X plus the distance from xn to Y where here I'm applying symmetry so that I can move the terms around right and this we know that for all Epsilon bigger than zero there exists an N let's say this is equal to the maximum of N1 and N2 in the natural numbers such that both of these terms are now less than Epsilon in fact you know what to make it nicer I'll make this Epsilon over two both spots so then this is less than Epsilon / 2 plus Epsilon / 2 which is equal to Epsilon now does this imp immediately that the uh that the distance between them is zero essentially we have to note one more thing which is that distances are positive definite so this is going to be bigger than or equal to zero so what this tells you is that because these the distance between X and Y gets arbitrarily small and arbitrarily small and close to zero it has to be zero this is from the real number theory that we've already know so this implies that the distance from X to y is zero and therefore X is equal to Y which is what we've wanted to prove so notice that a lot of this used all the aoms that we were talking about in one point we use symmetry which wasn't too important but was useful to note so that we can swap X and and X around and apply the triangle inequality we apply the triangle inequality and we had to use positive definiteness so you know these are the Bare Bones of what we really need in the theory all right so now what we're going to do is show that not only do limit points exist nicely but the distances between limit points act nicely so to state that more clearly theorem let y be an X and xn converge to X then my claim is that the distance from xn to Y will converge from the distance from X to Y which makes relative sense but notice here the statement is slightly different before we were purely dealing with sequences of a metric space and so we had to deal with the metric D itself but when we're saying that this converges to dxy what space am I considering these distances in no no no worries it's the we're looking at it in R right because for every single n this is just a real number and so when I say the distance from X and to Y converges to the distance from X to Y I mean that this happens in ukian space and in fact you can likely I don't want to State explicitly but I'm pretty certain you can prove this as well in a general setting so between two metric spaces okay so let's prove this to do this we want to show ultimately we want to show that for all Epsilon bigger than zero there exists an n and the natural numbers such that the distance from X and to Y minus the distance from X to Y is less than Epsilon and absolute values and to do so we can just find an upper and lower bound on the distance from X to y or the distance from X and to Y so let's do that the first Direction is nice the distance from X to Y is less than or equal to by the triangle inequality the distance from xn to X plus the distance from X to Y and we can make this term arbitrarily small so uh I'm just going to choose the same end to be the one for this convergent sequence so this is less than Epsilon plus the distance from X to Y so this gives us the upper bound that we're wanting on the distance from X to Y let's prove the lower lower bound and the lower bound is very similar but it's just a slightly different manipulation where what we're then going to do so the distance from xn to Y uh actually you know I'll look at the distance from X to Y this is less than or equal to the distance from X to xn plus the distance from X to Y and then what we can do is subtract these terms so specifically we'll get that the distance from X to Y minus the distance from X to xn xn is less than or equal to the distance from X to Y and we can make this term arbitrarily small so this is less than the distance from X to Y minus Epsilon and notice that this implies the result because then we have that the distance from xn to Y is less oh shoot did I do I get this wrong uh no no okay yeah so then the distance from xn to Y is less than sorry this should be the distance from X to Y there we go the distance from xn to let me double check this distance from xn to Y plus xn 2 y there we go sorry about that I I mixed up what they were converging to so what this tells us is that the distance from X into y minus the distance from X to Y gets arbitrarily small right I can write that out if y'all would prefer uh but that's just the next step you write it in terms of the absolute values okay any questions so not only do limit points be not only are limit points unique but the distances between limit points are also going to be unique at least in the real number sense which is pretty nice uh I'll remark one proposition which is pretty interesting you can study this for two convergent sequences at once so suppose or I I'll State this in two parts suppose xn converges to X and Y N converges to Y then the distance from xn to YN converges to the distance from X to Y this relatively makes sense but we can also State this slightly differently or a different type of this theorem suppose xn and YN are koshy so they're koshy sequences so they get arbitrarily closed together but don't necessarily have a limit point then we know that the distance from xn to YN converges notice that I'm not saying that I'm not saying what it converges to I don't know that Koshi sequences have limit points in my metric space uh and I'll get back to that in a moment but nonetheless the sequence itself in the real numbers will converge and that uses the fact that the real numbers is Koshi complete now I'm not going to prove these two facts these are facts that are uh on your problem set but I do want to point out because this happened a lot last year you cannot assume that the limit points in the second part exist you can know that this implication implies the top one by uniqueness because convergent sequences are koshy as we're going to prove right now um so if you prove the second one first then you can make a small remark you have to explain why the limit point is the distance from X to Y but that's not too bad so let's prove that theorem uh is that the next theorem I wanted to say uh yeah I'll do that one now Koshi sequences oh sorry convergent sequences sequences are kosi sequences proof suppose xn is the convergent sequence we're considering and that xn converges to X then we know that and I'm going to State this over and over again just like we do in real analysis but at a certain point it just BEC secondhand nature uh so therefore for all Epson bigger than zero there exists n in the natural numbers such that uh the uh such that for all n bigger than or equal to n the distance from xn to X is less than Epsilon so what does this tell us well what we're ultimately interested in is the distance from xn to XM right let M be bigger than or equal to n what we're interested in is the distance from X and to XM to state that it's a Koshi sequence so what can we do anyone sort of want yall to think about it because there's we want to bound this and show that this is less than Epsilon exactly the triangle inequality this is less than or equal to the distance from xn to X plus the distance from X or X to XM and we can make both of these less than Epsilon now if I was really careful I could have made these epsilons over two and then this would have been exactly Epsilon most of the time if you show it's less than a constant time Epsilon you're good you would just have to relabel things so all good things here but yeah exactly you apply the triangle inequality and notice oh and then this is precisely what we want for a Koshi sequence right this implies that it is in fact a koshy sequence so we're done uh now the real question is if koshy sequences are convergent sequences and as we know from the real numbers this is not always or from the real numbers we know that there it's Koshi complete but here we don't and what do I mean by Koshi complete which I've stated a few times definition a space is Koshi complete if and only if I.E the definition uh the Koshi sequences are convergent so this is the definition of what Koshi complete actually means and this is a very powerful fact right we use Co completeness everywhere in real analysis to state that limit points exist to state that differentiability was nice things like that we we needed Koshi completeness there now not every space is Koshi complete unfortunately but the ones that we really want them to be are in fact so uh proposition this is on your homework we have that yeah the set of continuous functions on 0 to one is Koshi complete this is a statement on your homework that I've broken up into individual steps so they should be a little bit nicer um but yeah a very useful proposition to show okay now let me go back to what I meant to do slightly earlier okay we have a few more things to say about uh Co convergent sequences and to do so I'm going to Define what it means for a set to be bounded right when we were looking at real analysis one of the most useful theorems we had in our tool belt was the bosano virus drust theorem now we're not going to have an analog of it but we still have some statements akin to it that we want to show uh and this will be especially important as we build up to compact sets so definition a sequence xn is bounded by let's say B bigger than zero if for all n in the natural numbers uh oh sorry I should say if there exists a point p in your metrics base such that for all n and the natural numbers the distance from xn to P is less than b which is what we would expect we want the convergent we want the sequence itself to to be the distance between the point and your sequence to be bounded that's what it means for a sequence to be bounded and if you want to picture this uh you can I always like to draw little pictures in my notes about this if this is our metric base X here's our Point p and what this is stating is that there exists a radius large enough of radius R sorry not R of radius B such that it completely contains X now granted what does it mean to have stuff outside of X nothing really so really we just mean the region in the middle is contained in the ball of radius r or radius B but that's a nice picture if you prefer and not only can we say this about convergent sequences or sorry not only can we say this about sequences in general we can also Define what this means for a set a set a in X is bounded by B if for all p in X oh sorry if there exists a p in X such that for all a and a the distance from a to P is less than b does the uh set theory notation that I'm using make sense to everyone there exists in for all cool so this is what it means for a sequence and a set to be bounded why do I bring this up because convergent sequences are bounded sequences we have this from real analysis but I'm going to prove it now in metric baces so proposition or I should I'll just write it out how I've been writing out the other ones if xn converges to X then xn is a bounded sequence okay let's start off with the proof which we'll start off as all of our statements have so far with the statement of what uh of what convergence means and the statement of what convergence means I'm going to choose the Epsilon that I want so that I'm not dealing with all the possible epsilons so I'm going to note for Epsilon equal to one bigger than zero we know that there exists an n in the natural numbers such that the distance from such that for all n bigger than or equal to n the distance from xn to X is less than Epsilon which again I'm assuming is one now does this complete our statement the answer is no here we only have that this is true for all n bigger than or equal to capital N right but what we want for our statement is to to show that for every single natural number this sequence is bounded but this isn't an issue because what we can note is that there are finitely many terms less than n in the natural numbers right so what we can do is that b be the maximum of the finitely many terms of the distance from x i tox or let's say uh or one for I between 1 and capital N so what I'm doing here is I'm noting that here I have my convergent sequence I know that most of them in fact infinitely many of them are contained uh are contained in a ball of radius one and I only have finitely many terms outside of that ball so what I'm doing is I'm saying okay well it's either in this ball or the next one or the next one or so on and so forth this is what the statement is that we wanted to show and we know that this is in fact finite as there was only finitely many terms so this is the B that it's bounded by I can write that out some more if yall would prefer but this is essentially the statement cool uh so I think that's mostly oh there's one more thing we want to say about convergent sequences which is that their subsequences act nicely which will be essentially how we want it to be um so proposition let xn converge to X and xn K be a subsequence of xn then my claim is that xn K is convergent and in fact to show that it's convergent we're going to show that it converges to X right which is what we should expect of a subsequence does everyone here know what a subsequence is happy to redefine um cool um so let's go with the statement so proof notice that my subsequence here is arbitrary I'm not going to rewrite that out but we could choose any subsequence of our sequence and what we're interested in is we want to show for all Epsilon bigger than zero there existant n in the natural numbers such that for all n k now bigger than or equal to n the distance from X and K to X is less than an Epsilon and now again any guess this as to what we should do triangle inequality precisely yeah so we're considering the distance from X and K to x what we can do is apply the fact that we know that convergent sequences are koshy so we can State we can write this as less than or equal to the distance from xn K let's say to XM for m bigger than or equal to n plus the distance from XM to X and now I can choose a larger natural number if necessary such that this term gets less than Epsilon / 2 and this term because it's Koshi is less than Epsilon over 2 so without writing out all those steps this is less than Epsilon so this shows that the distance from X and K to X gets arbitrarily small so therefore it's convergent um yeah this is fairly important and it highlights you know one of the many ways in which Koshi sequences are deeply important as we'll talk about in like the fifth lecture in a specific module of this class uh having things to be Koshi complete really really helpful um okay oh one more note this is something that I vaguely noted on uh Tuesday but these are all of the major theorems and propositions that we're going to need for convergent sequences you might be wondering why there aren't more we had a ton more in real analysis in fact it took like a month of our time the reason we don't have more is because the real numbers are a vector space I can add two real numbers and get a real number I can multiply them and get a real number everything there was nicer and we had addition toin so when it meant so one thing we could state in real analysis is that this for instance the sum of two convergent sequences is convergent here that doesn't make sense necessarily you can't always add two points in a metric space and get a point in your metric space let alone have addition be well defined uh similarly we have the squeeze theorem in real analysis we have a version of it based off of distances but we don't have a squeeze theorem for points so because we don't necessarily have an ordering on our set right you can picture the complex numbers the complex numbers don't have a uh ordering on them you don't have a sense in which one complex number is bigger than the other other than yeah I'm just going to leave it there if you've done complex analysis then you'll know what I'm talking about uh but yeah so that's why we don't have more theorems as we used to so now we're going to move on to the next major theorem that I've talked about or the next major definition that I've talked about we've talked about convergent sequences we've talked a little bit about koshy sequences kosy sequences are helpful but in terms of a metric space we've mostly St stated what we need for the moment being uh now we're going to move on to open sets so I'm going to recall what this definition is because it's the one that can be a little bit the weirdest so recall a set a contained in your metric Space X is open if for all for every single point in a there exists an Epsilon bigger than zero such that the ball of radius Epsilon around a is completely contained in your set and this is the set of Y in your metric space such that the distance from y to a is less than Epsilon uh just to bring it up because sometimes the notation comes up this is also sometimes denoted the ball around a of radius Epsilon so use whichever notation you prefer uh can you here I can move it a little bit up so use whichever notation you prefer uh I prefer this one just because it means uh less commas but you know whichever okay so uh open sets are going to have a huge connection between topology and continuity and other definitions that are important now the first thing that we're going to prove is topological properties of open sets so because youall haven't taken topology this is in fact really really helpful we're going to show three major properties of uh open sets in metric spaces so a theorem uh and you can call this topological properties if you want or I'll write it out cu topological properties of open sets Okay firstly we have that the empty set and the entire metric bace itself are open two given AI are open sets then the union of all of these sets from I equals let's say one to infinity or all see Zero why not uh is open so in other words the arbitrary Union of open sets is open and finally uh the finite intersections of open sets are open in other words if I intersect from IAL 0 to capital M of a I then this is open the fact that we can't intersect infinitely many of them will become apparent in the proof so in the proof of these three properties it'll become apparent why we can't do infinitely many of them but let's see this one so consider the empty set how do we know that the empty set is open well it's vasly true it's true all the time because for every single point in the empty set there are no points in the empty set so as soon as that statement is false we have that the empty set is open so this is open vake is sleeve two let's consider the set X well for all Epsilon or sorry for every single point x and x does there exist a ball around x and x yes there is because just pick pick your favorite Epsilon bigger than zero and then the ball of radius Epsilon around that point x is by definition a subset of your metric space right because the definition of a ball of radius Epsilon is only the point in X so we're not going to end up with some weird thing outside of X so this shows that X itself is open two we want to show that the union of open sets is open how' we do that um IAL 0 to Infinity is open this is what we want to show well as as with the definition we're just going to pick a point in the union so pick arbitrary X in the union I equal 0 to Infinity of a i and we want to show that there exists a ball of radius Epsilon around X that's contained in the set well because it's in the union there has to exist at least one AI that contains X so therefore there exists an AJ in your set of open sets AI such that X is in AJ and then we apply the fact that AJ itself is open so therefore there exists an Epsilon bigger than zero such that the ball of radius Epsilon around X is contained in AI or sorry AJ but this is a subset of the Union so this shows that the ball of radius Epsilon is contained in the infinite Union or potentially infinite Union this also works if it's finite notice but you know that that's just a subcase okay uh and can I squeeze it in here no I cannot so I will move over here any questions so far actually let me move with this any questions [Applause] open sets are weird in that you know when we learn about them in real analysis I feel like they're pretty unintuitively important everyone says that they're important but I never got why and I'll explain why they are just like in a moment um but yeah open sets are deeply important in real analysis Sy topology and we're going to prove some more statements besides these topological properties that will be important uh and we'll relate to the major definitions in our tool tool box okay so now we're interested in the third case the intersection from I equals 0 to M of the AI well we're going to do exactly as we did before if I have some X the intersection of the AI from i = 0 to M then what do I know before we knew that it was in any of the AIS but here we know even more information here we know that X is in AI for every single I so what can we do well we know that therefore because each AI is open there exists an Epsilon I bigger than zero such that the ball of radius Epsilon I around X is contained in AI the issue is that the entirety of AI might not be in the intersection so what can we do we have this for every single I and we have finitely many of them so choose uh Epsilon to be equal to the minimum of the Epsilon I from I equals z or 0 to m capital M this is bigger than zero right because we're only choosing finitely many this is where we're using the fact that there's the finite intersection if we had an infinite intersection you can picture taking um more and more uh smaller and smaller Epsilon I such that this uh infinum is zero and that's an issue but if it's finitely many we can take the minimum of them and then what we know is that the ball of radius Epsilon round X is a subset of AI for all I because it's contained in the ball of radius Epsilon I I guess I should tell that this is contained in the ball of radius Epsilon I of X which is contained in AI for all I which implies that the ball of radius Epsilon around X is a subset of the finite intersection all right all right so why is this called the topological properties of open sets because if you were to take 18901 which is a class that I would highly suggest to take at some point if you're interested in this sort of analysis uh as we're going to see open sets allow us to state things about continuity you can Define continuity in terms of open sets you can Define convergent sequences in terms of open sets basically everything we can redo in terms of open sets and that is a topology if I have a set of subsets such that these three properties hold that's known as a topology and in 18901 that's the basic toolbox that you're given and then you go from there in terms of redefining everything so this is just an example of uh a more General thing than metric spaces um I'm going to talk a little bit about that in the final lecture where I talk about you know where things go from here but for now this is you know the basic idea all right uh uh now what I'm going to state is some facts about closed sets but what is a closed set which I haven't defined before so this is useful to note a subset a in X is closed if a uh if the complement of a which is known as this the metric bace x minus a is open now if a set is closed does that imply that it's not open yeah the answer is no how can we see that well let's consider the real numbers as an example uh notice what's the complement of the empty set in the real numbers anyone it's a good exercise and thinking about what the complement means real numbers minus the empty set so it's just going to be all of the real numbers but this as we've stated before in our first topological property this is open so therefore the empty set is closed by definition but we also know by our topological properties again that the empty set is open what the heck well this is just one of those things in topology it's kind of weird like open sets you can prove properties of closed sets instead and things are everything everything's fine um I'll note note there's a notion called connectedness which I'll briefly Define But ultimately will be more of a problem set problem but if your metric space is connected uh let me state it that way if if your metric Space X is connected which I'll redefine in a mo or which I'll Define in a moment this is true if and only if the only open and closed sets so sets that are both open and closed are the empty set and the metric space itself now what does it mean for your metric space to be connected I mean we have a nice picture of it in our heads CU connected is a pretty visual thing but this is one definition once you prove this this is a definition how do you prove this well you prove it based off of the definition of disconnected so definition X is disconnected by definition if there exists two open sets U1 and U2 that are joint disjoint and non-empty such that uh the union of them is exactly X this is what it means for your space to be disconnected and from here a set is connected if it is not disconnected so kind of kind kind of circular reasoning and then once you have that you can prove this note which I'll put in the problem or the third problem set which is an interesting problem um but I just want to knowe showing something is closed does not necessarily imply that it's open uh I'll give one short example of why this is true and then I'll move on to more properties of open sets anyone have questions cool uh feel free to shout out questions if you have any or interrupt I'm more than happy to be interrupted okay so I'm just going to give one example of a metric space that is disconnected which is one that makes relative sense at least on the face of it uh example I can take the a union of two open intervals 0 to 1 and 1 to two and give it the usual metric on R what do I mean by usual metric I mean the distance between two points in this set is just the distance normally in R right absolute values this is an example of one that's disconnected right because it's the union of two or one that is is disconnected it is the union of two open sets right that are disjointed and you can notice here the interval 01 is both open and closed so uh because the complement of it is just one two and that's open so this is an interesting example to think through this is known as the Subspace metric it's really or the Subspace topology technically um but yeah okay let's go back to proving things about the general theory of open sets proposition I'm first going to point out that closed sets are very similar to top uh very similar to open sets in fact we can Define everything in terms of of closed sets if you wanted to properties of closed sets here we have that the empty set and X are closed to the infinite the potentially infinite intersection of closed sets AI closed is closed and three the finite Union of Clos sets is closed nearly identical to the topological properties of open sets how do we prove this it's not too bad I'm not going to actually do it the fact that the empty set and X are closed is just the proof that I did above for R right you can replace r with just any metric space uh and we have that this first property holds how about the other two well for the other two I'm just going to quickly note what are known as the Morgan's laws so this is is a nice Lemma from met from set theory these are known as de Morgan's laws going to switch which talk I'm using this states that the complement of a union is the intersection of the complements so I and a and similarly the complement of an intersection UI complement I in a is equal to the union of the complements these are known as de Morgan's laws they make relative sense you can prove them for two sets as opposed to infinitely many of them but once you have these two properties then you can rewrite these two properties in terms of open open sets that's why there's this Duality right I can write the intersection of closed sets as the union of open sets so that's uh that's how you apply to Morgan's laws and I'm not going to prove that or I'm not going to prove these topological properties this is in the lecture notes but I do want to knowe you could do everything in terms of closed sets if you really truly wanted to we just don't because open sets are kind of nice okay we'll show why some of those things are actually nice uh what more things do I want to know okay cool uh now I want to State some specific examples of open sets and Metric baces which we can already do quite a bit of example uh one thing did I erase it already I think I oh wait no I did it here uh I did it where did I do it oh I did it at the very top uh the ball of radius Epsilon around a is sometimes referred to as an open ball but you should prove that the ball is in fact open which will be our first example so given some point x in your metric Space X and Epsilon bigger than zero we have that the ball of radius Epsilon round X is open how do we do this well we choose an arbitrary point in your ball of radius Epsilon right so proof uh choose some y in the ball of radius Epsilon around X we want to find a ball of radius let's say Delta around y contained in the ball of radius Epsilon around X that's the definition of open uh open set well notice that the distance from X to Y is less than Epsilon by definition right because we're assuming that the distance between points in our ball is less than Epsilon so let Delta be the distance from X to Y over two why does this work because then we'll notice that the ball of radius Delta around y has to be contained in the ball of radius Epsilon around X why is this true essentially the triangle inequality right we have a ball of radius Epsilon centered at X have a point Y here and I'm noting that the distance from here to here is Epsilon uh is less than Epsilon by definition and I'm just choosing a ball around y to be half of that distance and then you can apply the triangle inequality to show that this ball must actually be contained in the bigger one anyone want me to work through that I'm happy to oh wait one let me let me write it out uh you know I'll write it out on the backboard so let z b in the ball of radius Delta around y then we have that the distance from X to Z is less than equal to by the triangle inequality the distance from X to Y plus the distance from y to Z uh distance from X to Y and the distance from y to Z is less than the distance from X to Y over 2 is this what I wanted to do epon Prim uh I should say no this is not what I wanted to do let let R be the distance from X to Y minus Epsilon then what we know is that the distance from y to Z is less than R what am I trying to say distance of Y to Z is less than or sorry this should be Epsilon minus the distance there we go and this is then less than Epsilon so this is in the lecture notes I would highly suggest drawing out the picture and working through it uh separately but Loosely speaking it's just this diagram on the right hand side it's an application of the triangle inequality okay so the ball of radius Epsilon is an open ball in fact you can State much more than this this is an optional problem on your second problem set and for the record I do suggest yall look at the um look at the optional problems because they will potentially give you more intuition even if you don't solve them um but yeah it's this is just one of those cases where this isn't a topology class so I'm not going to ask that y'all do a bunch of topology but it's pretty nice okay so just a small note any open set U contained in your metric space can be written as a union a potentially infinite Union of open balls I'll I'll outline briefly how to do this right now so if youall want to do it later you can do it um but theide idea is that for any single point in your open set there exists a ball of radius Epsilon around that point and then you can just take the union of all those balls of radius Epsilon the Epsilon might change for every single point but that doesn't matter right because it's an infinite Union so I'll let youall work through those details but this is just a proposition you can show okay let's also just prove what an example of a Clos set which will come important for next time it's just an example x uh sorry I should say let X be a point in your metric space then the set X is closed so this is just a singular point in your metric space how we're going to prove this well we only have one definition of closed we have that it's the complement or the complement is open right so uh proof consider the complement of X which is X not including X and let y be a point here what we want to show is that there's a ball of radius Epsilon I want to show around y or not in subset of X not including X let's draw a picture of what this should actually look like right uh well I didn't want to do that here's our metric space here's our singular point x and here's our Point let's say y we want to choose a ball around y that doesn't contain X how can we do it anyone you could choose whatever Epsilon you want so it's a question of what Epsilon want to choose we can choose Epsilon to be the half the distance between X and Y I think that's what I was mixing up on the other proof but it's in the lecture notes so I'll let you all do the other one but um choose Epsilon to be the distance from X to Y over two and then pictorially we have that it's you know that's rest of the problem but let's actually work through the details so then let Z be in your ball of radius Epsilon around y then we want to show that Z cannot be X right well notice this implies that the distance from y to Z has to be less than Epsilon right this is by construction but what can we can we get a lower bound on this we can we can note that the distance from I want to make sure I'm getting this right r x actually you know I'll do this directly or I'll do this by contradiction suppose for the sake of contradiction X was in the ball of radius Epsilon around y then this would imply that the distance from X to Y is less than Epsilon but this can't be the case because again we're assuming that the distance that Epsilon is the distance from X to Y over 2 and the distance from X to Y cannot be less than the distance from X to Y over 2 so that is our contradiction I'll make one more small note here which will come important for Tuesday which is as opposed to looking at a singular point you can do this with a finite Union of points right and the argument would be essentially the same as our finite intersection of open sets argument you have a finite Union of open sets or sorry a finite Union of closed things so it should be closed in fact you could use that to prove it immediately but you can also prove it directly like this okay so now what we're going to do is talk about how this definition of open set that's relates to convergent sequences and to um and to continuity proposition let xn to do this I'm going to do it in the real numbers but note that everything can be done in your metric bace I just want to use the notation from the real numbers let xn be a sequence am I holding it backwards I am in R uh that converges to X so this is true I claim if and only if for all Epsilon bigger than zero all but finitely many x i are in x - Epsilon to X Plus Epsilon so this is the ball of radius Epsilon around X right how do we prove this if and only if proof well okay I should say convergent sequence let's prove the forward Direction first so if xn is convergent then for all Epsilon bigger than zero there exists an n in the natural numbers such that the ball of radius Epsilon around x uh such that I should say such that for all n bigger than or equal to n the uh xn is in the ball of radius Epsilon X Plus Epsilon as this is true because the distance from xn to X is less than Epsilon this is just the open set way to view it right and this capital N notice that there are only Finly many terms less than it so this shows that infinitely many of the terms are in this open set and in fact only finitely many are outside of it so that proves the forward Direction let's prove the opposite direction uh suppose or I should write suppose uh for all Epsilon bigger than zero all but finitely many many XI are in x - Epsilon to X Plus Epsilon I want to choose a sequence that will converge to X then right well how can we do that we can construct a subsequence that is going to be convergent and if it's remember a subsequence if it's convergent converges to what the original sequence converges to right so this is where we're using a fact from earlier that we proved okay so suppose Epsilon is 1/ n for n and the natural numbers or I'll say m then choose X n m in the ball of radius Epsilon so 1 / M to x + 1 M what does this tell us I claim that this convergent sequence converges to X well to show that we need to write down the definition when professor riguez was erasing the board he'd always tell some joke but I don't have any jokes right now unfortunately though he would do them two at a time so he could actually have time to actually say the joke before he finished okay so we want to show that X and M converges to X right proof for all Epsilon bigger than zero we can choose M uh large enough or I should say capital M such that for all n m bigger than or equal to M we have that uh 1 / n m is less than Epsilon right because epsilon's bigger than zero I can find a large enough natural number such that one over that natural number is less than Epsilon that's our it's a factor from real analysis so what this tells us is that we know then that X and M or I should say uh yeah X and M is in xus epsilon to x + Epsilon because this is from X over 1 n m to x + 1 n m which is a subset of the B radius Epsilon by construction right because 1 nm is less than Epsilon so what this tells you is that the distance from xn M - x is less than Epsilon which implies that the convergent sequence or the convergent subsequence converges to X so to reiterate how the proof went we had a convergent sequence we had a subsequence that we showed converges to x and by what we showed earlier X must be the limit point of the original convergent sequence then so we're done all right uh I won't write out the analogous statement for metric bases the you just change xus Epsilon to X Plus Epsilon with a ball of radius Epsilon um I just dated it in real real numbers because the I prefer the interval notation to think about it and it's a fact that like not everyone fully sees in real analysis because sometimes you don't know what an open set is in 100a it depends on who's teaching it realistically okay we're going to prove one more oh wait what do I want to say yeah okay this was just why I wanted to show that it's related to uh convergent sequences now what I'm going to do is show that it's related to continuous functions and I'm going to recall what that definition is because that can again be a little bit weird between metric spaces so a function f from a metric Space X to a metric space y let's say with metric DX and Dy is continuous if for all Epsilon bigger than zero there exists a Delta bigger than zero such that if the distance between two points X and Y is less than Delta then the distance in the metric space y of the images of X and Y is less than Epsilon this is the the definition of continuous in the metric bace setting and so now what we're going to do is prove a statement akin to convergent sequences but for continuity we can write the definition of continuity in terms of open sets and not only can we do that we can write continuity in terms of convergence of sequences so I'll do that right now theorem under all the same assumptions here I'm going to let X and Y be metric spaces with distances DX and Dy and yeah okay uh let claim uh F from X to Y is continuous at a point C in X if if and only if if xn is a sequence that converges to X then F of xn converges to F ofx this is my statement uh when I say continuous at the point C I mean just replace uh y with C here that means that this is continuous at every Point continuous at x and x and as continuous if it's continuous everywhere right this is the same as in real analysis we had get continuity at a point and continuity everywhere this is just continuity at a point okay proof I always prefer going the downwards Direction first so let F be continuous at C then by definition for all Epsilon bigger than zero there exists a Delta such that this implication holds what does this actually tell us well suppose we're going I'm going toose suppose the um uh thing in the second hypothesis suppose xn is a sequence in X that converges to X then what can I say well we know that there exists a sorry I should say to C then by definition of convergent sequence we know that for all Epsilon bigger than zero there exists an n in the natural numbers such that for all n bigger than or equal to n the distance from xn to C is less than Epsilon but we know that for all Epsilon bigger than zero furthermore there exists a Delta bigger than zero such that the distance from F of xn to F of C is less than Delta over two oh sorry not Delta over two just Delta uh is this no uh I should have re labeled this I should have said for all Delta bigger than zero there exists one such that this is less than Delta and then there exists an Epsilon bigger than zero such that no no I said it right the first time uh for all bigger than Z there exists a Delta bigger than zero such that this term is less than Delta and furthermore the distance between the images is less than Epsilon there we go that's what I meant to say and so what this tells you is that F of xn must converge to F of C right because there exists an n in the natural numbers such that the distance between the two points is less than Epsilon which is the definition of convergent sequence and then to prove the other direction uh I'm just going to State it because are running out of time no I'll write it out I'll write it out over here to proof the other direction we're going to use it by uh contradiction we're going to assume that it's not continuous at C and then choose a convergent sequence that uh or a sequence that doesn't converge or such that the images don't converge okay if and only if suppose f is not continuous this is the proof of the upwards Direction suppose it's not continuous at C what does this tell you well saying something is not continuous is a little bit weird but I'll State out what that definition means given some bigger than zero we know that there exists there exists an xn in the sequence such that the distance from xn to C get small uh I want to say one over n I'll just say less than given Epsilon bigger than Z there exists a Delta bigger than zero and an X in your sequence such that the distances get small so let's say 1/ n but the distances between the images of them F of C is bigger than Epsilon right because the the uh Contra uh the opposite would be if this was less than or equal to Epsilon right and so then what you can show is that therefore xn converges to C but F of xn does not converge to F of C we know that it doesn't converge to F of C because we just proved up here that if a sequence is to Converge All But finitely many of them I.E infinitely many of them or that's not quite the same but all but finitely many of them must exist in any ball of radius Epsilon right and we just showed here that there are in fact infinitely many that don't exist in a ball of radius Epsilon so uh that proves the other direction this is known as sequential continuity sometimes if you want to study uh this in more generality as opposed to assuming continuous you can assume this this sort of definition on topological space and that's slightly different um but for metric spaces the two definitions are the same right which will be the case a lot of the time a lot of the time in this class A definition that's specific to metric spaces will imply other very useful definitions that might not hold in more generality okay I just have one more statement to show which is how continuity relates to open sets uh let me write out what this Sumo will be oh sorry I should have this is poor board manners I should have raised this in the with this okay okay just one more Lemma f is continuous at C if and only if oh I need to state a definition first I'll state it over here definition a neighborhood of a point Y is simply an open set that that contains y containing y for our purposes though you can just think of this open neighborhood as a ball right because I've stated that every uh open set can be written as a union of finitely of infinitely many balls so this is what a definition of a neighborhood is and that lets us State the Lemma which is for every open neighborhood neighborhood of Y oh sorry of f of C this will be our y then we have that the inverse image F inverse of f of oh sorry I should say every open neighborhood U of f of c f inverse of U Is An Open neighborhood of f of C sorry of C because we're in the inverse image okay has anyone seen this sort of definition before regarding open sets yeah um it doesn't come up all the time because it is essentially topology like if you take a topology class you would see this definition of this would be the definition of continuity in a topological class um but here we're going to actually prove it for a metric bace and because not all of you have seen it before we'll just go through that proof so proof we'll prove the downwards Direction so we have that f is continuous at C and I'm going to let you be an open neighborhood of f of c and so then I want to show that F inverse of C oh sorry F inverse of U Is An Open set of C open neighborhood of C right what I'm going to do instead uh instead of considering this entire open neighborhood I'm going to consider a ball of radius Epsilon right around F of C so we know because U is open we know that there exists an Epsilon bigger than zero such that the ball of radius Epsilon