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5. Metric Space Topology

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You have (presumably) been learning about metric space topology by defining what open sets of the metric space \((X, d)\) are. In fact, one may develop the theory of topology from an axiomatic point of view! A topology on a set (not necessarily a metric space!) is simply a description of which subsets are open, subject to a handful of conditions.

We can drive our intuition of what open sets are by thinking about \(\mathbb{R}^n\) under the usual Euclidean metric. In this setting, open sets are just sets that “have open boundaries”. Recall that closed sets are complements of open sets; continuing this analogy, closed sets are those that “have closed boundaries”.

One may very loosely think of topologies as describing how points in a set are organised, and open sets describe which points are “close together”. This notion of closeness is quite imprecise and purely qualitative, and it’s made quantitative through defining a metric.

There are two main goals for today’s discussion.

First, going off of this loose and informal intuition, some features of metric space seem to be purely topological while others are not. For instance, distances and size are quantitative and are not topological. On the other hand, convergence is qualitative: \(x_n\to x\) in a metric space if \(x_n\) gets “arbitrarily close” to \(x\). So, our goal will be to distinguish which features of metric spaces are actually topological features, to be made more precise later.

Second, although our mental imagery of \(\mathbb{R}^n\) is a good source of intuition, there are lots of very unintuitive topological facts that can become sources of mathematical error later on. We’ll identify a few of these unintuitive facts that you better be aware of.

Definitions and Techniques

Before jumping in, let’s refresh our memories of the important definitions, then discuss a few techniques for proving topological statements in metric spaces.

While you often won’t be asked to show that a specific set is open or closed, the open-ness or closed-ness of sets often form important criteria for theorems later on. Moreover, open sets and closed sets often satisfy “nice” properties that are related to the metric (such as the open ball criterion in the very definition of open-ness)! So how does one verify that a set is open or closed?

For open-ness, it’s often easiest to proceed directly from the definition and show that given a point in your (hopefully open) set, you can find some ball around that point contained in said set.

Showing closed-ness is more difficult. From the definitions, to show that \(S\subseteq X\) is closed, one must verify that \(X\setminus S\) is open. But sometimes this is not a very accessible definition: for certain metric spaces and choices of \(S\), the complement can be extremely difficult to describe and work with. Moreover, the sets that we care about are often sets of elements satisfying nice properties (such as continuity within sets of functions); elements outside these “nice” sets are difficult and should be subject to capital punishment.

Here is one particularly helpful characterisation of closed sets in metric spaces:

Try proving this yourself before looking at the proof! It’s a good exercise, and it’s an important fact too.

Proof

First, suppose \(C\) is closed. Let \(\left\lbrace x_n \right\rbrace\) be a sequence in \(C\), and let \(L\) be its limit; we wish to show that \(L\in C\).

Suppose towards a contradiction that \(L\notin C\). Then \(L\in X\setminus C\), and since \(X\setminus C\) is open, there exists some \(\epsilon>0\) such that \(B(L, \epsilon)\subseteq X\setminus C\). However, by the definition of convergence, there exists some \(n_\epsilon\) so that for every \(n>n _{\epsilon}\), \(d\left( x_n,L \right)<\epsilon\). In particular, \(x_n\in B(L, \epsilon)\) for all \(n>n_\epsilon\). This is impossible since \(x_n\in C\) while \(B(L, \epsilon)\subseteq X\setminus C\).

Now suppose \(C\) contains all of its limits; we wish to show that \(C\) is closed. We will instead show that \(X\setminus C\) is open. Suppose towards a contradiction that \(X\setminus C\) is not open. There exists some \(y\in X\setminus C\) such that for all \(\epsilon>0\), \(B(y, \epsilon)\) is not contained in \(X\setminus C\). Equivalently, for all \(\epsilon>0\), the intersection \(B(y, \epsilon)\cap C\neq \varnothing\).

Define a sequence in \(C\) as follows: for any \(n\in \mathbb{N}\), pick some \(x_n\in B\left( y,\frac{1}{n} \right)\cap C\). Such an \(x_n\) exists because we just showed that this intersection was nonempty. Then \(x_n\) is a sequence in \(C\), and \(x_n\) converges to \(y\)! For any \(\epsilon>0\), there is some \(n_\epsilon>0\) such that \(\frac{1}{n_\epsilon} <\epsilon\). Then, for all \(n>n_\epsilon\), we have \[d\left( x_n,y \right) < \frac{1}{n} < \frac{1}{n_\epsilon} < \epsilon.\] The leftmost inequality is because \(x_n\in B\left( y,\frac{1}{n} \right)\). Thus, \(x_n\to y\) by definition. However this means \(y\in C\) since we assumed \(C\) contains all of its limits, and this contradicts \(y\notin C\)! We conclude \(X\setminus C\) must be open, hence \(C\) is closed. \(\square\)

I should mention that this proposition does not say that every sequence in a closed set does converge to some limit in that set. Instead, it says that if the sequence converges to begin with, its limit must be in the closed set. This is a very important distinction!

We have thus identified an important technique.

• To show that a set \(C\) is closed, start with a sequence \(\left\lbrace x_n \right\rbrace\) in \(C\) that converges to some sequence, possibly outside of \(C\). Then, show that this limit must actually be in \(C\).
• To show that a set \(S\) is not closed, construct a sequence of elements \(x_n\in S\) that converge to some limit that in fact does not lie in \(S\).

In fact, we can take this one step farther. We defined the closure of a set (S) to be the intersection of all closed sets containing (S). This is a difficult definition to work with; barring degenerate cases, it’s hard to even describe all of these closed sets to begin with. Instead, we offer an explicit characterisation:

Proof

This proof will happen in three steps:

1. \(C\) contains \(S\).
2. \(C\) is closed.
3. Every closed set containing \(S\) also contains \(C\).

For the first step, if \(s\in S\), define the sequence \(x_n=s\) for all \(n\). Then clearly \(\left\lbrace x_n \right\rbrace\subseteq S\) and \(x_n\to s\), hence \(s\in C\) by construction.

For the second step, \(C\) is closed. Let \(\left\lbrace c_n \right\rbrace\) be a sequence of elements in \(C\) converging to some \(L\in X\). We claim that \(L\in C\). For each \(n\), since \(c_n\) is the limit of a sequence of elements in \(S\), there exists some \(s_n\in S\) such that \(d \left( c_n,s_n \right) < \frac{1}{n} \) for all \(n\). Then, \[d\left( s_n,L \right)\leq d\left( s_n, c_n \right)+ d\left( c_n, L \right)\leq d\left( c_n,L \right)+ \frac{1}{n}.\] Given \(\epsilon>0\), let \(N_1>0\) such that \(n>N_1\) implies \(d\left( c_n,L \right)< \frac{\epsilon}{2} \), and let \(N_2>0\) such that \(n>N_2\) implies \(\frac{1}{n} < \frac{\epsilon}{2} \). Then, for all \(n>\max\left( N_1,N_2 \right)\), we use \[d\left( s_n,L \right)< \frac{1}{n} +d\left( c_n,L \right)<\epsilon\] to conclude that \(s_n\to L\). Thus \(L\in C\) by construction. By the previous proposition, since \(C\) contains all of its own limits, \(C\) is closed.

Finally, if \(C’\) is any closed set containing \(S\), by the proposition again \(C\subseteq C’\) — every sequence in \(S\) is a sequence in \(C’\), and \(C’\) must contain all of its limits. Thus \(C\subseteq \overline S\) and \(\overline S\subseteq C\), hence \(C=\overline S\). \(\square\)

This allows us to very simply describe closures of sets in metric spaces.

Topological vs. Metric Properties

Recall that every set \(X\) may admit many different metrics. What we care about the most in a metric space is the convergence of sequences, and as mentioned before, this is a topological feature. We’ll make this more rigorous in the following problem:

In fact, this says that if two metrics produce the same open sets (i.e., they create the same topology), then they produce the same convergent sequences and limits. Conversely, if two metrics produce the same convergent sequences and limits, then they induce the same topology on the set! This is what we mean when we say “convergence is a topological feature” — it is solely determined by the topology associated to a metric, not by the metric itself. On the other hand, some quantitative features of metric spaces aren’t actually topological.

There is a picture for what’s going on. \(\arctan\) bijectively maps \(\mathbb{R}\) to the open interval \((-\pi,\pi)\) — it “squashes” the real line into a smaller space, and the metric \(d’\) reflects this “squashing”. Nonetheless, this finite interval and the whole real line are topologically indistuingishable in the sense that they have the same “shape”.

Topological Bear Traps

When I was first learning about open and closed sets, I thought that these produced a dichotomy of subsets, i.e. that every set was either open or closed. But this is not quite true — one can have sets that are neither closed nor open, and one can even have sets that are simultaneously open and closed.

By the way, the topology induced by the trivial metric is called the discrete topology. This tells you no information about how you should organise or draw your set, as the topology says that your set is merely a “discrete” collection of points, nothing more and nothing less.

As a fun remark, what you have just shown is that \(\mathbb{Q}\) with the topology from the usual metric makes \(\mathbb{Q}\) a “totally disconnected space”, and this reflects the fact that \(\mathbb{Q}\) is permeated by a lot of “holes” where real numbers should be.

Finally, recall the notation of open balls, closed balls, and closures of sets. We should point out that the closure of an open ball is not necessarily the closed ball of the same radius. Symbolically, this means that \[\overline{B}(x, r)\neq\overline{B(x, r)}\] in general, where \(x\) is a point in a metric space and \(r\geq 0\) is the radius of the balls.