Supplementary definitions: levels of abstraction

Although we have been able to generalise a lot of our theorems about the topological properties of metric spaces to topological spaces, several others remain out of our reach, and we need to find the right level of abstraction that makes them true -- and we can usually do this by trying to prove the theorem and seeing what it "needs", then if this theorem actually characterises the space, proving the need is equivalent to the theorem.

Separation axioms

We've seen the first one. TFAE:

• $X$ is a T0/Kolmogorov/distinguishable space.
• All points in $X$ are distinguishable.
• For any pair of points, there exists an open set containing exactly one of them.

Now, in metric spaces, statements like "there is a point of a set (or sequence) $S$ in every open neighbourhood of $x$" are equivalent to "there are an infinite number of points of $S$ in every open neighbourhood of $x$", i.e. every open neighbourhood of a limit point of $S$ has infinite points in $S$. The key here is that at any step in the process, an open neighbourhood of $x$ can be constructed that does not contain any of the previous points in the sequence, i.e. there exists an open neighbourhood of $x$ not containing some finite number of points.

This is possible, e.g. if we decide that finite sets are closed (or equivalently singletons are closed). This is an iff -- if a finite set $S$ isn't closed, there are points in $\mathrm{cl}(S)$ that aren't in $S$, but it's impossible for there to be an infinite number of points in $S$ in a neighbourhood of such a point, as $S$ is finite.

Wait, what about for points in $S$ itself -- aren't these all also limit points of $S$? We sneakily changed the definition of limit points from $x\in\mathrm{cl}(S)$ to $x\in\mathrm{cl}(S-\{x\})$, i.e. to exclude sequences that include the point itself. So since excluding a set from a finite set keeps it finite, it is still closed, and thus $x$ is not re-included when you close it (so it is not a limit point at all).

Wait, what about the discrete topology on a finite set? All finite sets are closed, but how could we have an infinite number of anything? The key is that there are no "limit points" of any set. So the only T1 finite spaces are the discrete ones.

Note how this implies a stronger form of T0 -- T0 said that for every pair of points, at least one of them has an open neighbourhood excluding the other. Well, in a T1 space, we can always just remove any finite number of points from an open neighbourhood and it will remain an open neighbourhood -- so in particular, this means that for every pair of points, each has an open neighbourhood excluding the other, or any two points are separated. In fact, this is clearly an iff (take some finite intersections).

The closedness of finite sets has plenty of other implications that match our intuition. You will see some in the list of exercises.

So for now, TFAE:

• $X$ is a T1/Frechet space.
• All points in $X$ are separated.
• For any pair of points, each has an open neighbourhood excluding the other.
• Each open neighbourhood of a limit point of $S\subseteq X$ contains an infinite number of members of $S$.
• Finite sets are closed/singletons are closed.

We've seen spaces in which limits are not unique. You might think that topological distinguishability should suffice to have unique limits -- is this right?

Let's go through the proof of the uniqueness of limits on metric spaces. Suppose $f$ has two limits $L_1$ and $L_2$ at $c$. Now does it really suffice that there is a neighbourhood of $L_1$ not containing $L_2$? Not necessarily -- it may be the case that this neighbourhood of $L_1$ intersects every neighbourhood of $L_2$ because there's just that amount of indiscretion around $L_2$. What's important is that there exist disjoint neighbourhoods of $L_1$ and $L_2$.

So the uniqueness of limits is equivalent to the existence of disjoint neighbourhoods for distinct points. This is variously called a Hausdorff space or a T2 space. So TFAE:

• $X$ is a T2/Hausdorff space.
• All points in $X$ are separated by neighbourhoods.
• Any pair of points has a pair of respective neighbourhoods disjoint from each other.
• Limits of nets/filters are unique.

Here's another obvious fact about the topology of metric spaces: any continuous function $f:S\to\mathbb{R}$ on a closed (not just compact) set $S$ may be extended to a continuous function $F:X\to\mathbb{R}$. The proof of this relies on the ability to "connect" points between the pieces of $S$:

You might think that this just means "having a continuous function between two points", but this is only because of the choice of a one-dimensional space -- in general, the "boundary" of each connected component is not just a finite number of points. What does suffice is that for disjoint closed sets $S$ and $T$ there exists a continuous function $h:X\to\mathbb{R}$ "separating" them, i.e. such that $h(S)=\{0\}$, $h(T)=\{1\}$. Then a bunch of such functions can just be added to $f$ to obtain $F$. This is trivally an iff.

One may try further to "prove" the existence of such a continuous function $h$. Well, on $\mathbb{R}$, we have a notion of a "midpoint" between the endpoints of the closed sets -- we could have the value of $h$ at this point be $1/2$, and then continue the construction for all "Dyadic rational"-points.

In general, we don't really need the notion of a midpoint, but we do need some point with "space around it", i.e. open neighbourhoods of each closed set that don't contain the point. We only need the case where $S$ and $T$ are in a connected component, so this is equivalent to the existence of open neighbourhoods of disjoint closed sets. This is also clearly an iff, as one may consider preimages of open sets in $[0,1]$. Thus TFAE:

• $X$ is a T4/normal space.
• Every pair of disjoint closed sets has respective disjoint open neighbourhoods. 
• Every pair of disjoint closed sets can be separated by a continuous function (Urysohn's lemma).
• Every continuous map on a closed subset can be extended to a continuous map on $X$ (Tietze's extension theorem).

Countability

We've already seen the first one. TFAE:

• $X$ is a first-countable space.
• Every point in $X$ has a countable neighbourhood basis.
• $\lim_{x\to a}f(x)=L\iff\forall (x_{n\in\mathbb{N}})\to a, f(x_n)\to f(a)$. 
• Every filter has a corresponding sequence.

Several other cardinality functions can be defined and other axioms of countability can be introduced, e.g. sequantial spaces, separable spaces and second-countable spaces, but we cannot really motivate them at this point. (Second-countability in particular important when dealing with metrisability, integration and related issues, as it gives rise to the notion of "paracompactness".)

For the following statements, find the "least restrictive" level of abstraction necessary, and decide if the property characterises the level (i.e. is it an iff?):

1. Every subset $S$ is the union of all the closed sets contained in $S$.
2. Every subset $S$ is the intersection of all the open sets containing $S$.
3. Every nonempty open set contains a nonempty closed set.
4. Every point $x$ admits a nested neighbourhood basis, i.e. a neighbourhood basis totally ordered by $\subseteq$.
5. Closed subset $T$ of a compact space $S$ is compact.
6. Compact space is closed.
7. There is no smallest neighbourhood of any point (i.e. $\forall x\in X,\forall N\in N(x), \exists M\in N(x), N\not\subseteq M$).

Solutions:

1. T1. This is equivalent to asking that every point in $S$ is contained in a closed subset of $S$, which is possible if the singleton is closed. For equivalence, consider a singleton $S$.
2. T1. Dual to Exercise 1.
3. T1 suffices, but is not equivalent (e.g. the indiscrete topology or any other topology full of clopen sets). Maybe if you change to proper containment.
4. First-countability. (both sides of equivalence are clear -- think directed set, etc.)
5. Any topological space suffices. We need to be careful with the proof here because the notions of closedness and compactness are intuitively very similar. Here it is: any net on $T$ is a net on $S$ and thus has a convergent subnet with limit is in $S$ -- but its limit must also be in $T$ because $T$ is closed.
6. Hausdorff suffices (probably not equivalent, but who cares). The thing is that in general, even on a compact space, although a convergent sequence may have a convergent subsequence in $S$, it may have other limits outside $S$. But in a Hausdorff space, limits are unique.
7. "Not Alexandrov". Let's characterise the topologies that do have smallest neighbourhoods -- these are necessarily the ones for which arbitrary intersections of open sets are open, and are called finitely-generated topologies or Alexandrov topologies. 

Topology II: Kuratowski closure topology, nets, neighbourhood basis

So far, we've described two axiomatisations of toplogy: in terms of neighbourhoods and in terms of open sets. While the neighbourhoods definition was a natural extension of our understanding of topological structure being in terms of limits, the open sets definition is kind of hard to wrap your head around, as far as I can see. A continuous function doesn't even preserve open sets (the definition of a continuous function isn't "preserves neighbourhoods" in the neighbourhood formulation either, but at least we know where it comes from). It's not openness in particular that's important, we could formalise topology in terms of their complements the closed sets, too.

In the following set of exercises, we will build an alternative axiomatisation of topology based on the notion of touching, which will perhaps give us some explanation of why the notion of openness and closedness are important to topology.

1. Consider the relation of "touching" between a point and a set (it can't be between a point and a point, but it can be with a set -- kind of for the same reason that a sequence tends to a point but there is no point in the sequence that equals that point). How would you write this in terms of the open set topology? (Ans: every open set containing $x$ intersects $S$)
2. Now formulate an open set in terms of touching. (Hint: formulate a closed set first, the open set is its complement) (Ans: no point in $S$ touches $S'$)
3. Great. Find out what axioms we need on the touching operation $\sim$ to prove the three axioms of open sets.
1. $X$ is open requires -- (Ans: $\not\exists x \sim \varnothing$)
2. $O_1\cap O_2\in\Phi$ requires -- (Ans: $x\sim S\cup T\Rightarrow x\sim S \lor x\sim T$)
3. $\bigcup_\lambda O_\lambda \in\Phi$ requires -- (Ans: $x\sim S\subseteq T \Rightarrow x\sim T$)
4. Given a touching relation $\sim$, we can produce the set of open sets $\{S\mid \forall x\in S, x\not\sim S'\}$. From this, we can produce the relation $\bar\sim$, given by: $\forall T\, \mathrm{st.} (x\in T \land \forall y\in T, y\not\sim T'), S\cap T\ne\varnothing$. Try proving this is equivalent to $x\sim S$, and see what axioms you need.
1. $\Leftarrow$ requires -- (Ans: None. Suppose $S\cap T$ were empty. Then $S\subseteq T'$, so by 3c-ans, $x\sim T'$, contradiction.)
2. $\Rightarrow$ requires -- (Hint: >given such a $T$, construct a smaller $T$ whose intersection with $S$ is empty if $x\not\sim S$) (Ans: $S\subseteq\mathrm{cl}(S)$ and $\mathrm{cl}(\mathrm{cl}(S))\subseteq\mathrm{cl}(S)$. Suppose $x\not\sim S$. Then for any $T$ satisfying the LHS (e.g. the universe by 3a-ans), consider $T\cap\mathrm{cl}(S)'$ where $\mathrm{cl}(S)$ is the set of all points touching $S$ -- $x\in T\cap \mathrm{cl}(S)'$ by assumption; now given $y\in T\cap \mathrm{cl}(S)'$ we want to show $y\not\sim T'\cup \mathrm{cl}(S)$ (for this to contradict the claim that $S\cap(T\cap\mathrm{cl}(S)')\ne\varnothing$, we need that $S\subseteq\mathrm{cl}(S)$ -- this is new!). Suppose that $y\sim T'\cup\mathrm{cl}(S)$ and apply 3b-ans: $y\sim T'$ contradicts $y\in T$ as $T$ is open; now we just want $y\sim\mathrm{cl}(S)$ to imply $y\in\mathrm{cl}(S)$ -- this is new!)
5. So rewrite our axioms in terms of the closure operator $\mathrm{cl}$ as follows -- and this is completely equivalent to our earlier "touching" description of the closure operator as $x\sim S\iff x\in\mathrm{cl}(S)$:
1. $\mathrm{cl}(\varnothing)=\varnothing$
2. $\mathrm{cl}(S\cup T)\subseteq\mathrm{cl}(S)\cup\mathrm{cl}(T)$
3. $S\subseteq T\Rightarrow \mathrm{cl}(S)\subseteq\mathrm{cl}(T)$
4. $S\subseteq\mathrm{cl}(S)$
5. $\mathrm{cl}(\mathrm{cl}(S))\subseteq\mathrm{cl}(S)$
6. To get yourself comfortable with this closure operator, check if the following statements are true or false:
1. $\mathrm{cl}(S)\cap\mathrm{cl}(T)\subseteq\mathrm{cl}(S\cap T)$ (Ans: False -- consider two disjoint sets sharing a boundary)
2. $\mathrm{cl}(S\cup T)=\mathrm{cl}(S)\cup\mathrm{cl}(T)$ (Ans: True -- in fact can replace 5b and 5c)
3. $\mathrm{cl}(\mathrm{cl}(S))=\mathrm{cl}(S)$ (Ans: True -- it's also equivalent to the statement that $\mathrm{cl}(S)$ is closed, i.e. that no point in $\mathrm{cl}(S)'$ touches $\mathrm{cl}(S)$ -- can you see why?)
4. $S\subseteq\mathrm{cl}(T)\Rightarrow \mathrm{cl}(S)\subseteq\mathrm{cl}(T)$ (Ans: True -- in fact can replace 5c and 5e)
5. $\mathrm{cl}(S\cap T)\subseteq\mathrm{cl}(S)\cap\mathrm{cl}(T)$ (Ans: True -- in fact the arbitrary-intersection version of this is the reason we have the equivalent 5c, from 3c-ans, and can replace it)
7. From 5c and the answer to 6a, you might see an analogy with limits of sequences -- indeed, a "set" is kind of a sequence, and its closure is to add the limit points of the sequence to the set. Indeed, the definition of a continuous function, as we will see, is $f(\mathrm{cl}(S))\subseteq\mathrm{cl}(f(S))$, i.e. all limit points remain limit points (and for a homeomorphism, the reverse inclusion is also true). This definition should be fairly obvious as it just says "if $x$ touches $S$, $f(x)$ touches $f(S)$", i.e. nothing gets ripped apart in the process. So $\mathrm{cl}$ is a generalisation of a limit to any subset of $X$.
8. Well, actually, this is not clearly a stronger condition than "convergent sequence limits are preserved" -- that actually requires that convergent sequences remain convergent, that you don't just add a new limit point like you can for non-convergent sequences.

But this does get us thinking -- we saw above that "preserves the limit points of every set" fully describes a continuous function. But does "preserves the limits of convergent sequences" -- which we used to motivate the idea of topology in the first place -- still characterise a continuous function? Is $x_n\to a\Rightarrow f(x_n)\to f(a)$ (for all sequences $(x_n)$) equivalent to $x\to a \Rightarrow f(x)\to f(a)$ like it does for standard metric spaces? Certainly the backward implication is correct.

Let's go through the proof of the forward implication on metric spaces.

Suppose $f$ is not continuous at $a$. So $\exists\varepsilon>0$ such that $\forall\delta>0,\exists x\,\mathrm{st.} |x-a|<\delta, |f(x)-f(a)|\ge\varepsilon$. We want to construct a sequence $(x_n)$ converging to $a$ so that $\forall N, \exists k \ge N, |f(x_k)-f(a)|\ge\varepsilon$. We construct the $n$th element of the sequence from the choice function for $\delta = 1/n$, which is an $x$ within $1/n$ of $a$ such that $|f(x)-f(a)|\ge\varepsilon$. 

OK -- how can we generalise this to an arbitrary topological space? 

Suppose $f$ is not continuous at $a$. So $\exists N\in N(f(a))$ such that $f^{-1}(N)\notin N(a)$, i.e. $\forall M\in N(a), \exists x\in M, f(x)\notin N$ (make sure you can tell that these are indeed equivalent). We want to construct a sequence $(x_n)$ converging to $a$ so that $\forall K, \exists k\ge K, f(x_k)\notin N$. Now here's the deal: if we can construct a sequence of $M$s in $N(a)$ that ultimately converge to the point $a$, like we could in the metric case, we are done.

What does it mean to "ultimately converge to the point $a$"? We need that the choices are eventually contained in every neighbourhood of $a$ -- or to write it down concisely: we want a sequence of sets $M_i\in N(a)$ such that $\forall N\in N(a), \exists i, M_i\subseteq N$. This is called a neighbourhood basis -- and particularly since the domain of $i$ is the natural numbers, a countable neighbourhood basis.

Well, so does every topology admit a countable neighbourhood basis to every point? As it turns out, there are counter-examples.

So we've generalised a bit beyond the notion of preserving limits of sequences, and this is OK -- the natural numbers aren't that fundamental, are they? I would say that preserving the limit points of sets as in pt. 7 is a more fundamental notion than preserving the limits of convergent sequences. In any case, different terms exist for different levels of specialisation, such as:

• Pre-topological space (where you can have "multiple layers of boundaries" -- this is what happens if you leave out idempotence of closure, the arbitrary union axiom of open sets, or the wiggle-room axiom of neighbourhoods)
• Topological space
• T0-space (distinguishability)
• T1-space
• T2-space or Hausdorff space (uniqueness of limits of nets)
• Alexandrov topology (with arbitrary intersection and union)
• First-countable space (for which the "limits of sequences" thing suffices)
• Second-countable space
• Separable space
• Uniform space
• Metrisable space

In any case -- although we do not always have a countable neighbourhood basis, we do always have a neighbourhood basis for a point -- the entire neighbourhood filter itself. And while the neighbourhood filter isn't a countable set, it is a directed set (a poset where every two elements have a shared superior). The generalisation of a sequence to a directed domain is called a net. 

We can define the limit of a net from a directed set $D$ in the same way as usual with filters: $\forall N\in N(a), f^{-1}(N)\in N(+\infty)$, or $\forall N\in N(a), \exists K\in D, \forall k \ge K, x_k\in N$. Then our generalisation of the "limits of a sequence" motivation for topology is:

A map is continuous if it preserves the limits of all convergent nets.

Of course, the convergence of nets is equivalent to the convergence of filters.

---

Dense sets

Let's discuss a quick application of closure -- recall how $\mathbb{Q}$ is dense in $\mathbb{R}$. This can be formulated in numerous ways, but the simplest way is probably "every open set in $\mathbb{R}$ intersects $\mathbb{Q}$. Does this remind us of something? Yes, of course -- it's the definition of closure in terms of open sets, i.e. $\mathrm{cl}(\mathbb{Q})=\mathbb{R}$.

And this is obviously true -- it's the definition of $\mathbb{R}$, isn't it? So there's a very natural explanation for the rationals being dense in the reals.