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.

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