# Discrete space $(X, \tau)$ is homeomorphic to a subspace of $\mathbb{R}$ iff $X$ is countable

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?

• As it's written right now the question is not understandable (for me), could you clarify what you're asking? – Alessandro Codenotti Jan 6 '17 at 18:20
• if $B \subset \mathbb{R}$ with Euclidean distance be homeomorphic with (X,T) that T is discrete topology then prove X is countable . – amir bahadory Jan 6 '17 at 18:23
• Are you asking "Prove that every discrete subspace of $\mathbb{R}$ is countable"? – Dan Rust Jan 6 '17 at 18:25
• 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. – Jorge Fernández Hidalgo Jan 6 '17 at 18:28
• @AlessandroCodenotti. yes. – amir bahadory Jan 6 '17 at 18:28

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.