Cardinality of a metric space

Question

Let $X$ be an infinite set .For any two metrics $d_1,d_2 $ on $X$ the identity map $i:(X,d_1)\to (X,d_2)$ is continuous. Prove that $X$ is always countable.

I am not getting how to start this problem. Please help.

Answer 1

Suppose we have such a set $X$, which is infinite. Let $A = \{a_n: n \in \mathbb{N}_0\}$ be a countable subset. Let $d_1$ be the metric that is $1$ between all distinct points of $X$, except for $d_1(a_i, a_j) = |\frac{1}{i} - \frac{1}{j}|$ for $i,j \ge 1$, $d_1(a_0,a_i) = \frac{1}{i}$. So $A$ behaves like an isometric copy of the sequence $\frac{1}{n}$ with its limit $0$, the rest is discrete. Set $d_2$ to be the discrete metric. But then the identity is not continuous, as $a_n \rightarrow a_0$ but this does not hold for the same sequence in the discrete metric $d_2$. So the assumption is void, no such $X$ can exist.

If $X$ is finite, then all metrics are discrete metrics (because then a minimal non-zero distance between distinct points of $X$ exists, as a finite minimum, and then all balls with radius smaller than that are singletons, so the topology is alway discrete), and the identity map is a homeomorphism indeed.