Problem of Separable Metric Space, Isolated points and countable sets

Question

How to prove that if a metric space $(E, d)$ is separable and $A\subseteq E$ is a set where all points are isolated then A is countable.

Answer 1

Definition. Let $(E,d)$ be a metric space.

• A base of open sets for $E$ is a family $\mathscr{B}$ of open subsets of $E$ such that, for any open subset $O$ of $E$, there is collection $\{O_i\}_{i\in I}$ contained in $\mathscr{B}$ such that $$O=\bigcup_{i\in I} O_i.$$
• If $F\subseteq E$, when we say that $F$ is a subspace of $E$, we are considering the metric space $(F,\rho)$, where $$\rho=d\big|_{F\times F}.$$

It is a good exercise to prove (one direction is provided in a Brian's comment) that:

Theorem 1. Let $E$ be a metric space. $E$ is separable if and only if $E$ has a countable base of open sets.

From this follows:

Corollary 2. Let $E$ be a metric space. $E$ is separable if and only if all its subspaces are separable.

Proof. Assume first that $E$ is separable. By Theorem 1, there exist a countable base of open sets $\mathscr{B}$. Let $F$ a subspace of $E$. Then the family $\mathscr{B}'=\{O\cap F : O\in \mathscr{B}\}$ is countable base of open sets for $F$. By Theorem 1 we conclude $F$ is separable.

Conversely, suppose that $E$ is such that all its subspaces are separable. $E$ is a subspace of $E$, so $E$ is separable.

From Theorem 1 we also get:

Lema 3. Let $E$ be a separable metric space. If $\{U_i\}_{i\in I}$ is a family of pairwise disjoint, open, nonempty sets, then $I$ is countable.

Proof. Since $E$ is separable we can assume that there is a base of open sets $\mathscr{B}=\{O_j\}_{j\in\Delta}$ with $\Delta\subset \Bbb N$.

Since the open sets in $\{U_i\}_{i\in I}$ are pairwise disjoint and nonempty, for each $i\in I$, you can pick (axiom of choice possibly needed here) an element $x_i\in U_i$. Notice that the $\{x_i\}_{i\in I}$ are necessarily pairwise distinct.

Because $\mathscr{B}$ is a base of open sets, for each $i\in I$ there exist a $j_i\in\Delta$ such that $$x_i\in O_{j_i}\subseteq U_i.$$

Notice that $i\neq k$ implies $j_i\neq j_k$. Otherwise there would be a pair of open sets in $\{U_i\}_{i\in I}$ whose intersection is not empty.

Then, the function $f:I\to\Delta$ given by $$f(i)=j_i$$ is an injection from $I$ to $\Delta$, a countable set. Therefore $I$ must be countable.

Now, the answer to your question:

Theorem 4. Let $E$ be a metric space. If $E$ is separable and $A$ is a subset of $E$ such that all its points are isolated, then $A$ is countable.

Proof. Consider $A$ as subspace of $E$. By Corollary 2, $A$ is separable.

Since all the points of $A$ are isolated, for each $x\in A$ there exist a $r_x\gt 0$ such that $$B(x,r_x)\cap A=\{x\}.$$ Thus $$\{B(x,r_x)\cap A\}_{x\in A}$$ is a family of pairwise disjoint open subsets of $A$. Since $A$ is separable, by Lemma 3, this family must be countable. Since there is exactly one set in that family for each element of $A$,$A$ must be countable.

Finally, it is worth to point out that

Corollary 5. Let $E$ be a metric space. If there exist a uncountable subset $A$ of $E$ such that all its points are isolated then $E$ is not separable

Answer 2

Every separable metric space is second countable, meaning that it has a countable base $\mathscr{B}$ for the topology. Suppose that $A$ is a subset of $E$ in which every point is isolated. Then for each $x\in A$ there is a $B_x\in\mathscr{B}$ such that $B_x\cap A=\{x\}$. The map $A\to\mathscr{B}:x\mapsto B_x$ is injective ($1$-$1$), so ... ?

Answer 3

Assume $A$ is uncountable. Since a point $a\in A$ is isolated form other points in $A$, there is an open ball $U_a$ of radius $r/2$ about $a$ where $0<r< \text{min}_{b\in A, b\neq a}d(a,b)$. Check that $U_a\cap U_b = \phi$. Then the open subspcae $\cup_{a\in A}U_a$ of $A$ is not seperable, it has no countably dense subset. This makes a contradiction.

Answer 4

In fact, every CCC (Countable Chain Condition space) has countable isolated points. Note that every separable space is CCC. Refer Here.

Proof: Suppose not. Let $E=\{x_\xi: \xi\in \omega_1\}$, where $x_\xi$ is an isolated for each $\xi \in \omega_1$. Then it has uncountable disjoint open sets $\{\{x_\xi\}: \xi\in \omega_1\}$, which leads a contradiction.

Answer 5

$(E,d)$ separable means there exist a countable dense subset $D=\{x_n:n\geq 1\}\subset E$.

Now $x \in A$ isolated means that there exist an $\epsilon$-ball centered at $x$ containing no points other than $x$. If $x \notin D$, then this ball contains no points of $D$ whatsoever, but this contradicts the density of $D$. Hence $x \in D$.

We have shown that $A\subset D$. Since $D$ is countable, it follows that $A$ is countable also.