Let $\mathbb{R}$ have the Euclidean topology (i.e., with Euclidean distance), and let $(X, \tau)$ be a discrete topological space.

Prove that $(X, \tau)$ is homeomorphic to a subspace of $\mathbb{R}$ iff $X$ is countable.

Let $X$ be countable. If $|X|$$=n$, then the subspace $A=\{1,2,3,...,n\}$ with the induced Euclidean topology is homeomorphic to $(X, \tau)$. If X is countably infinite, then the subspace $\mathbb{N} $ is homeomophic to $(X, \tau)$.

Now how we can prove the other direction, i.e., if $(X, \tau)$ is homeomorphic to a subspace of $\mathbb{R}$, then $X$ is countable?

  • $\begingroup$ As it's written right now the question is not understandable (for me), could you clarify what you're asking? $\endgroup$ Jan 6, 2017 at 18:20
  • $\begingroup$ if $B \subset \mathbb{R} $ with Euclidean distance be homeomorphic with (X,T) that T is discrete topology then prove X is countable . $\endgroup$ Jan 6, 2017 at 18:23
  • $\begingroup$ Are you asking "Prove that every discrete subspace of $\mathbb{R}$ is countable"? $\endgroup$
    – Dan Rust
    Jan 6, 2017 at 18:25
  • $\begingroup$ I think he means: prove that there is a subspace $A$ of $\mathbb R$ such that a subspace of $A$ is discrete if and only if it is countable. $\endgroup$
    – Asinomás
    Jan 6, 2017 at 18:28
  • $\begingroup$ @AlessandroCodenotti. yes. $\endgroup$ Jan 6, 2017 at 18:28

1 Answer 1


HINT: Prove the contrapositive: show that if $A\subseteq\Bbb R$ is uncountable, then the relative topology on $A$ is not discrete. One way to do this is to let $\mathscr{B}$ be a countable base for the topology of $\Bbb R$, and let

$$\mathscr{B}_0=\{B\in\mathscr{B}:B\cap A\text{ is countable}\}\;.$$

  • Show that $A\cap\bigcup\mathscr{B}_0$ is countable.

Let $A_0=A\setminus\bigcup\mathscr{B}_0$.

  • Show that $A\ne\varnothing$, and that each $x\in A_0$ is a limit point of $A$. Conclude that $A$ is not discrete in its relative topology.

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