# Showing that all non-empty compact, dense-in-themselves metric spaces are uncountable.

Given a metric space $$(X,d)$$ and its associated metric topology $$\tau$$, $$(X,d)$$ is dense-in-itself if and only if every non-empty open set in $$\tau$$ contains more than one element of $$X$$. Also, $$(X,\tau)$$ is compact if and only if every open cover of $$X$$ contains a finite subcover.

I would like to find a way to show that all such non-empty metric spaces are uncountable, preferably via a proof that uses the covering property explicitly and does not require handling sequential compactness or completeness—for example, showing that a countable dense-in-itself metric space must have an open cover that does not have a finite subcover.

• nit: Presumably you want to add "nonempty" to "every open set in $\tau$ contains more than one element of $X$." Jan 5, 2022 at 21:27
• Oh yes—thanks for the catch! Jan 5, 2022 at 21:32
• You didn't understand the comment of @JairTaylor. It was talking about "... if and only if every nonempty open set in $\tau$ contains more than one element of $X$". Also I don't know why you are hating sequential compactness and completeness. To me it seems that it's natural to prove this result by using completeness. Jan 5, 2022 at 21:34
• I see; made the correction. Re: sequential compactness and completeness, it is precisely because it seems natural to prove the result with completeness that I would like to see if it's possible to do so "tidily" without invoking it, or at least, without requiring an external definition of completeness. (Although it may possibly not be worth the effort...) Jan 5, 2022 at 22:18
• Up to homeomorphism $\Bbb Q$ is the only countable metric space without isolated points (open singletons), since your space is compact it cannot be $\Bbb Q$ hence it must be uncountable Jan 5, 2022 at 22:44

We first note that for any $$x \in X$$ and any $$\varepsilon > 0$$ the closed ball $$\bar{B}_\varepsilon(x) = \{ y \in X : d(x, y) \leq \varepsilon \}$$ contains at least two points. This is direct because it contains the nonempty open ball $$B_\varepsilon(x)$$. Point of terminology: I will only consider closed balls with nonzero radius, so every closed ball will contain at least two points.

By induction we will construct closed balls $$C_\sigma$$, indexed by finite sequences of $$0$$s and $$1$$s. That is, for each finite sequence $$\sigma$$ of $$0$$s and $$1$$s we will construct some closed ball $$C_\sigma$$. Furthermore, we will do this such that for any $$\sigma$$:

1. $$C_{\sigma \frown 0} \subseteq C_\sigma$$ and $$C_{\sigma \frown 1} \subseteq C_\sigma$$;
2. $$C_{\sigma \frown 0} \cap C_{\sigma \frown 1} = \emptyset$$.

Here $$\sigma \frown i$$ denotes the sequence $$\sigma$$ with $$i$$ appended to it. So you can picture this as an infinite binary tree of closed balls. The further you go down a branch, the smaller the balls get. And at every node, the two next balls are disjoint.

We can start by taking $$C_\emptyset$$ to be any closed ball (where $$\emptyset$$ is the empty sequence). Having constructed all $$C_\sigma$$ for sequences of length $$n$$ we are going to construct the next layer. Let $$\sigma$$ be any sequence of length $$n$$ and pick two distinct points $$x, y \in C_\sigma$$. Let $$\varepsilon > 0$$ be small enough such that $$\bar{B}_\varepsilon(x)$$ and $$\bar{B}_\varepsilon(y)$$ are contained in $$C_\sigma$$, but are also disjoint (exercise: show that we can do this). Set $$C_{\sigma \frown 0} = \bar{B}_\varepsilon(x)$$ and $$C_{\sigma \frown 1} = \bar{B}_\varepsilon(y)$$. We do this for each such $$\sigma$$, and this completes the induction.

We will now need an equivalent formulation of compactness, but its equivalence to the finite subcover formulation is really elementary (again: exercise).

A topological space is compact if whenever a family of closed sets has the finite intersection property (i.e. every finite subfamily has nonempty intersection) then the entire family has nonempty intersection.

In notation: if $$\{F_i\}_{i \in I}$$ is some family of closed sets such that for any finite $$I_0 \subseteq I$$ we have $$\bigcap_{i \in I_0} F_i \neq \emptyset$$ then $$\bigcap_{i \in I} F_i \neq \emptyset$$.

We will view an infinite sequence of $$0$$s and $$1$$s as a function $$f: \mathbb{N} \to \{0,1\}$$. We will write $$f \upharpoonright n$$ for the initial sequence of length $$n$$. Let $$f$$ be such an infinite sequence. Then we consider the family of closed sets $$\{C_{f \upharpoonright n} : n \in \mathbb{N}\}$$. This family has the finite intersection property. Indeed, by construction we have a decreasing chain $$C_{f \upharpoonright 0} \supseteq C_{f \upharpoonright 1} \supseteq \ldots$$, so any finite part will have exactly intersection $$C_{f \upharpoonright n}$$ for some $$n$$. By compactness we thus have that $$\bigcap \{C_{f \upharpoonright n} : n \in \mathbb{N}\}$$ is nonempty, and we may pick some $$x_f$$ in that set.

So now we have $$x_f$$ for every $$f: \mathbb{N} \to \{0,1\}$$, and of the latter there are uncountably many. All that remains to show is that this assignment is injective. That is, for $$f \neq g$$ we have $$x_f \neq x_g$$. Let $$n$$ be the first $$n$$ such that $$f(n) \neq g(n)$$. Then by point 2 in the inductive construction we have $$C_{f \upharpoonright n} \cap C_{g \upharpoonright n} = \emptyset$$. At the same time we have $$x_f \in C_{f \upharpoonright n}$$ and $$x_g \in C_{g \upharpoonright n}$$, which concludes our proof.

• Let $2^{[[n]]}$ be the set of binary sequences of length $n> 0.$ Let $C_s=\bar B_{r(s)}(y_s).$ Let $R_n =\max \{r(s):s\in 2^{[[n]]}\}.$ We can make the sequence $(R_n)_n$ decrease to $0,$ so if the metric is complete then any sequence $(y_{f|_n})_n$ is Cauchy, & converges to $x_f$ where $\{x_f\}=\cap_{n\in\Bbb N}C_{f|_n}.$ Thus it suffices that $(X,d)$ is a complete metric space but not necessarily compact. Jan 6, 2022 at 6:13

A countable metric space that is dense-in-itself is homeomorphic to $$\Bbb Q$$ (see a proof here); This is due to Sierpiński and classic. $$\Bbb Q$$ is not compact (not bounded so a cover without a finite subcover is trivial to find) so you're done.

• That's a quite funny result. It implies e.g. $\Bbb Q^2$ is homeomorphic to $\Bbb Q$. Jan 6, 2022 at 4:36
• @WhatsUp it is indeed. Zero-dimensional spaces often behave like that. The irrationals are also homeomorphicto their square, and the Cantor set too. Jan 6, 2022 at 8:31