# Gracious Living

Countability Axioms
December 31, 2010, 01:55
Filed under: Math, Topology | Tags: , , , ,

So far we’ve seen two basic families of properties of topological spaces.  The connectedness axioms tell us in what ways it is possible to break our space apart into pieces.  The compactness axioms tell us how bounded the space is.  What we’re going to look at today is a set of axioms that deal with cardinality.  It should be mentioned that topology, for the most part, doesn’t really care about large cardinals — at most, we’re dealing with $\aleph_1$, the cardinality of our favorite counterexample $\omega_1$, and $c$, the cardinality of the reals.  These are equal if we accept the continuum hypothesis, and in either case we often talk about them in terms of countable subsets — sequences and the like.  The reason that countability is so important is that the properties we’re about to study are typical of metric spaces, and metrizability is a central question of point-set topology.

Basically, what we want to constrain is the cardinality of bases.

• A space is first countable if every point has a countable basis of neighborhoods.  A space is second countable if it has a countable basis.

Obviously, second countability implies first countability, and the converse is not true.  Because I don’t want to use $\omega_1$ again, here’s a new counterexample: take an uncountable set $X$, pick a point $p\in X$, and topologize $X$ by saying that the open sets are just those that are either empty or contain $p$.  Then every point $q$ has a minimal neighborhood: the set $\{p,q\}$.  This is a basis of neighborhoods of cardinality $1$ — so yes, it’s countable!  But any basis for $X$ would need to include an open subset of each set $\{p,q\}$, and the only way to do this is to include all of those sets, of which there are an uncountable number.  So it’s not second countable. $\mathbb{R}$, and in fact any metrizable space, is first countable.  Under any metric, each point $x$ has the set of balls $B_{1/n}(x)$ as a basis of neighborhoods.  Remember that any countable product of metric spaces is also metrizable with a metric that gives the product topology, so even $\mathbb{R}^\omega$ is also first countable. $\mathbb{R}$ is even second countable — as basis, we can choose the set of open intervals with rational endpoints (the rationals being countable, and the set of pairs of rationals thus being countable).  A basis for the product topology is given by the set of products of open sets, with all but finitely many of them being the space itself.  So we can even find a countable basis for $\mathbb{R}^\omega$ — the set of products of finitely many open intervals with rational endpoints and infinitely many copies of $\mathbb{R}$.

But not even all metric spaces are second countable.  For instance, let’s put a metric on $\mathbb{R}^\omega$ as follows.  Recall we can define a standard bounded metric on $\mathbb{R}$ by $\overline{d}(x,y)=\max(|x-y|,1)$.  Then there’s a bounded metric on $\mathbb{R}^\omega$ given by $\overline{rho}(x,y)=\sup_{1\le i}\overline{d}(x_i,y_i)$.  We call this the uniform metric, and the topology it induces the uniform topology.  The ball of radius $r<1$ around the origin is the set of points for whom the absolute value in each coordinate is at most some $s: that is, $\bigcup_{s (see why it’s not just $(-r,r)^\omega$?).  This makes the uniform topology strictly finer than the product topology and coarser than the box topology, where we can have open sets like $(-1,1)\times (-1/2,1/2)\times (-1/3,1/3)$ that contain no uniform ball.

If we look at the subset of this space consisting of all ordered tuples of zeros and ones, we see that the distance between any distinct points is $1$ — so as a subspace, this has the discrete topology.  It also has cardinality $2^{\aleph_0}=c$, so it’s uncountable.  But any discrete subspace of a set must correspond to a set of basis elements such that each one intersects only one of the points.  In particular, the cardinality of a discrete subspace is at most the cardinality of a basis of the whole space.  So uniform $\mathbb{R}^\omega$ cannot be second countable!  (As a corollary, all but countably many points of any subspace of a second countable space are limit points — try to prove this yourself.)

On the other hand, any compact metric space is second countable.  This is because we can cover such a space by its set of open balls of radius $1/n$, and then by compactness, reduce this to a finite subcover.  If we do this for every $n$, then the union of all these sets is a countable cover.  It’s also a basis: given any open set and any point, we can find a $1/n$-ball around that point in that set for some $n$, and then it’s pretty easy to show that that $1/n$-ball contains some $1/2n$-ball around the point and in our countable cover.  Euclidean $\mathbb{R}$ shows that there are also non-compact second countable metric spaces.

(By the way, $\mathbb{R}^\omega$ with the box topology isn’t even first countable — any countable basis of neighborhoods can be reconfigured with successive intersections such that $B_1\supset B_2\supset \dotsb$, and we could then find an open set that’s smaller than $B_1$ in the first coordinate, smaller than both $B_1$ and $B_2$ in the second coordinate, and so on.  This set wouldn’t contain any $B_i$.  But we already knew that the box topology was non-metrizable.)

To generalize, countable product and subspaces of (first, second) countable spaces are also (first, second) countable.  All the arguments are above, pretty much, but if you’re feeling confused you can work through them.

So why are these axioms important?  When we talked about metric spaces, we proved the sequence lemma:

• Lemma.  If $A\subset X$ and a sequence of points $x_n\in A$ converges to $x$, then $x\in\overline{A}$.  If $f:X\rightarrow Y$ is continuous, then for every sequence $x_n\rightarrow x$ in $X$, $f(x_n)\rightarrow f(x)$ in $Y$ as well.  The converses hold if $X$ is metrizable.

The surprising thing is that this also holds when $X$ is just assumed to be first countable!  In fact, knowing the definition is really all we need to prove this, because everything we said about $B_{1/n}(x)$ can be replaced with the $n$th basis element $B_n$ around $x$.  I urge you to look at the proof again and make the necessary changes.

What this means is that in a first countable space, we can characterize the closure of a set as the set of limits of sequences in that set.  And we can characterize a continuous function from a first countable space as one that preserves the convergence of sequences.  If the space isn’t first countable, we can’t necessarily do this, and closure and continuity might look weirder.

There are a few other things called “countability axioms,” some of which involve relaxing the constraints of compactness to countable rather than finite levels, some of which involve the idea of a “dense subset.”  But in the interests of keeping this short, I will leave you with these two for now.

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