# How to prove that $\mathbb R$ with usual metric and $\mathbb R$ with the discrete metric are not homeomorphic.

I know that two metric spaces are homeomorphic if there is a function from one to another such that $f$ is continuous, one-to-one, onto, with continuous inverse $f^{-1}$. I know how to prove two metric spaces are homeomorphic.

I don't understand how to prove that two metric spaces are not homeomorphic. I have to prove using knowledge I have of open sets, closed sets, limit and continuity. I cannot use compactness, disconnectedness, etc.

• It is not sufficient to show that a continuous function is one-one, onto. Consider $f : [0, 1) \to S^1, f(t) = (\cos 2\pi t, \sin 2\pi t)$. Commented Sep 3, 2014 at 14:34
• As @AymanHourieh indicates, you must also show that $f^{-1}$, which exists because $f$ is a bijection, is continuous.
– MPW
Commented Sep 3, 2014 at 14:43

Let $f\colon \mathbb{R}\to\mathbb{R}_d$ be a homeomorphism, where the domain has the usual metric and the codomain the discrete metric.

The sequence $(1/n)_{n>0}$ converges to $0$ in $\mathbb{R}$ with the usual metric, so $(f(1/n))_{n>0}$ should converge to $f(0)$ in $\mathbb{R}_d$. But in a discrete space only the eventually constant sequences converge.

Every set in the discrete metric is open. But there exist non open sets in the usual metric.

HINT: Consider any point $x$; $\{x\}$ is an open set in the discrete metric, use the openness criteria.

If you need more:

Let $\mathbb R$ be the reals with the usual metric, $\mathbb R^\dagger$ with the discrete metric. If $\phi: \mathbb R \to \mathbb R^\dagger$ is a homeomorphism, then the preimage of $\{x\}$ is a single point (injectivity), and that single point must be an open set in the usual metric.

• It's enough to note that every singleton is open in $\mathbb R^\dagger$ but no singleton is open in $\mathbb R$. This is a topological property, so the spaces are not homeomorphic.
– MPW
Commented Sep 3, 2014 at 14:38
• I got the feeling that his/her strategy was supposed to avoid topological properties (compactness, connectedness, etc.) for pedagogical reasons, but of course you are completely correct. Commented Sep 3, 2014 at 14:40
• Ah, yes, I didn't read carefully. Thanks for pointing out that detail.
– MPW
Commented Sep 3, 2014 at 14:42

Suposse that $(\mathbb R,d_1)$ and $(\mathbb R,d_2)$ are homeomorphic ($d_1$ usual metric and $d_2$ discret metric), then there exists a homeomorphism $f:(\mathbb R,d_1)\rightarrow(\mathbb R,d_2)$. Let $y\in\mathbb R$, consider the open set $\{y\}$ in $(\mathbb R,d_2)$, as $f$ is bijective, we have that $f^{-1}(\{y\})=\{x\}$ is a open set in $(\mathbb R,d_1)$. Thus we have that $(\mathbb R,d_1)$ is a discret metric space, which contradicts that $d_1$ is the usual metric.

Consider connectedness. Alternatively, one space is separable, and the other is not. Alternatively, the convergent sequences in one space are eventually constant, and you can count how many there are.

• OP says they cannot use connectedness. Commented Sep 3, 2014 at 14:39