Let $(X,d)$ be a separable metric space. Prove that every subset $Y \subset X$ can have at most a countable amount of isolated points.

Attempt at proof: Let $Y$ be an arbitrary (non-empty) subset of $X$. Since $(X,d)$ is separable, there exists a countable dense subset $A \subset X$ such that $\overline{A} = X$. Since $A$ is dense in $X$, we have that $A \cap Y \neq \emptyset.$ For every isolated point $x \in A \cap Y$ there exists then a $\delta_x > 0$ such that $B(x,\delta_x) \cap Y = \left\{x\right\}$.

Now I don't know how to prove the isolated points are countable. Any help?

  • $\begingroup$ Your attempt is wrong. Let $Y=X_A$ then the rest is just not applicable. Eventhough chizhek seems to have a solution I will try to give you a easier answer which is similar to your attempt. $\endgroup$ – YannickSSE May 17 '16 at 16:17

Let $C$ be the set of isolated points of $Y$. We will define an injective map, from $C$ to $A$, the countable dense set in $X$. The existence of such a map implies that $C$ is countable.

In order to do so, recall that for all $x\in C$ there is an $r_x>0$ s.t. $B_{r_x}(x)\cap Y=\{x\}$. Now let $f(x)$ be any point in $B_{r_x/2}(x)\cap A$, which is non empty by the density of $A$.

Now we are left to show the injectivity of $f$, let $x,y\in C$ s.t. $f(x)=f(y)$. Without loss of generality $r_x\geq r_y$. Then $f(y)\in B_{r_x/2}(x)$ and at the same time $f(y)\in B_{r_y/2}(y)$, hence $y\in B_{r_x}(x)\cap Y=\{x\}$. This is $x=y$. Hence $f$ is injective.


For metric spaces separability (there exists a dense countable set) is equivalent to the second axiom of countability (there exists a countable basis of open sets). $~$(If you do not yet know a proof of this fact, look it up.)$~$ It follows that a subspace of a separable metric space is separable.

More generally, we claim that a suspace of a second-countable topological space is always second-countable. Indeed, let $X$ be a second-countable topological space and $A$ its subset. The open sets
of the subspace $A$ are the traces $U\cap A$ of the open sets $U$ of $X$. Since $X$ is separable, is has
a countable basis $\mathcal{B}$ of open sets, which means that $\mathcal{B}$ is a countable set of open sets of $X$ such that every open set $U$ of $X$ is the union of the open sets belonging to some subset $\mathcal{U}$ of $\mathcal{B}$. Then the open set $U\cap A$ of $A$ is the union of the traces $B\cap A$ of the open sets $B$ in $\mathcal{U}$. Since every open set of $A$ is of the form $U\cap A$ for some open set $U$ of $X$, it follows that the set of all traces $B\cap A$ of sets $B$ in $\mathcal{B}$ is a countable basis of open sets of the subspace $A$, and we proved our claim.

Now it is easy to prove the assertion in your question. Since the metric space $X$ is separable, the metric subspace $Y$ is separable. Isolated points of $Y$ are the isolated points of the subspace $Y$.
Let $Z$ be the set of all isolated points of $Y$. The metric subspace $Z$ of the separable metric space $Y$ is separable, thus it has a countable basis $\mathcal{W}$ of open sets. But $Z$ is topologically a discrete space, that is, every point $\{z\}$ in $Z$ is an open subset of $Z$; since $\{z\}$ is the union of some sets in $\mathcal{W}$,
and the only nonempty subset of $\{z\}$ is $\{z\}$ itself, this one-point open set of $Z$ in fact belongs to $\mathcal{W}$.
But this implies that $Z$ is countable because $\mathcal{W}$ is countable.

[Added later.]

This is a commentary on the short and sweet, purely metric proof in the answer by "user3808066". The point of my approach is that by proving certain theorems about metric spaces (such as "every metric space containing a countable set has a countable basis of open sets") we use certain types of metric-space ("epsilon-deltic") reasonings only once, namely in the proofs of those theorems. Once we have the theorems, we can obtain certain other results about metric spaces using the theorems, without ever actually involving the metric. Of course, each of these other results can be obtained by straightforward metric reasoning, but when you take a closer look at the metric-space proofs you will see certain reasoning patterns repeated over and over. Each purely metric-space proof will replay parts of proofs of the theorems mentioned above.

As an example, and as an illustration of what I am talking about, I am going to prove equivalence
of separability and second-countability for metric spaces.

Let $X$ be a metric space.

Suppose $X$ has a countable basis $\mathcal{B}$ of open sets; we can assume that all sets in $\mathcal{B}$ are nonempty, since if the empty set happens to be in $\mathcal{B}$ we simply discard it. Select a point in each set in $\mathcal{B}$ and let $A$ be the set of selected points. The set $A$ is certainly countable, and it is dense in $X$: every nonempty open set $U$ contains some set in $\mathcal{B}$ and hence contains a point in $A$.

Conversely, suppose $X$ has a dense countable subset $A$. Let $\mathcal{B}$ be the set of all open balls $B_{1/n}(a)$ with a center $a\in A$ and a radius $1/n$, where $n$ is a positive integer. The set $\mathcal{B}$ of open sets is countable; we claim that it is a basis of open sets. Let $U$ be any open set and $x\in U$; we have to prove that there exists in $\mathcal{B}$ a set that contains $x$ and is contained in $U$. For a large enough positive integer $n$ the open ball $B_{2/n}(x)$ is contained in $U$. The open ball $B_{1/n}(x)$ contains some point $a$
in $A$ because $A$ is dense. But then $x\in B_{1/n}(a)\subseteq B_{2/n}(x)\subseteq U$.


Notice the radius $2/n$ -- half-radius $1/n$ trick. It's use is endemic in dealings with metric spaces.

[Added still later.]

Remark. $~$The fact that for a metric space separability implies that it is second-countable (a.k.a. completely separable), and that second-countability of a metric space suffices, without ever again using metric, to prove that the set of all isolated points of any subset is countable, all this means that the latter is true for any second-countable topological space. When you, in your problem, replace "separable metric space" with the equivalent "second-countable metric space", the property you then have to prove is about the topology of the metric space, no longer about its metric.

Remark. $~$"Countable" means "finite or countably infinite". "Almost countable" therefore means the same as "countable". Never use "countable" when what you really mean is "countably infinite".

  • $\begingroup$ I understand what you are saying. What I think is there is a place for big Theorems with a lot of implications, and there is a place for small proofs with very little tools. The former helps to understand the big picture and the latter helps to get a grasp of what is happening in this particular case. I think it is good that both found their way to this page! $\endgroup$ – YannickSSE May 18 '16 at 7:58
  • 1
    $\begingroup$ @user3808066 - I agree, completely. When you analyze some problem on its own terms, you get to know its peculiarities, and when you are done with it, you feel you know it intimately. But it does no harm then to take a stroll in the vicinity of the problem, observe its surroundings, and after a while to look back at the problem, seeing it anew as a familiar feature naturally positioned in the wider landscape. $\endgroup$ – chizhek May 18 '16 at 8:22

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