Possible cardinalities for continuity of the identity map with respect to any metrics Suppose $X$ is a metric space such that the identity map $id:(X,d_{1}) \rightarrow (X,d_{2})$ is continuous w.r.t any metrics $d_{1}$ and $d_{2}$. What are the possible cardinalities of $X$?
I know that $X$ can be any finite set, for then any subset of $X$ is open, and thus the pullback of any open set in the codomain is open.
What I am not sure about, though, is whether or not $X$ can be countably infinite, or even uncountable. Please provide me with some insights.
 A: Here's an example using a countably infinite set. Let $X = \Bbb{N}_0$ and
\begin{align*}
d_1(m, n) &= \begin{cases}|m^{-1} - n^{-1}| & \text{if } m, n > 0 \\ m^{-1} & \text{if } n = 0, m > 0 \\ n^{-1} & \text{if } n = 0, m > 0 \\ 0 & \text{if }m = n = 0. \end{cases} \\
d_2(m, n) &= |(m+1)^{-1} - (n+1)^{-1}|
\end{align*}
Both metrics are inspired by bijections between $\{1/n : n \in \Bbb{N}_+\}$ and $\Bbb{N}_+$ and $\Bbb{N}_0$. In this case, $d_2$ is the result of bijectively mapping this set onto $\Bbb{N}_0$ (resulting in the discrete topology), while $d_1$ is the result of mapping the completion onto $\Bbb{N}_0$.
I claim that the identity may $(X, d_1) \to (X, d_2)$ is not continuous. More specifically, under $d_1$ the sequence $(n)_{n=1}^\infty$ converges to $0$, but under $d_2$ this is certainly not the case.
This establishes an example for one (and hence all) countably infinite sets. I further claim that any set $X$ containing a countably infinite set (i.e. any infinite set) will also admit a counterexample. For such $X$, we find a countably infinite subset $Y$, and let $\phi : Y \to \Bbb{N}_0$ be a bijection. Then we define
$$d'_i(x_1, x_2) = \begin{cases} 0 & \text{if } x_1 = x_2 \\
d_i(\phi(x_1), \phi(x_2)) & \text{if }x_1, x_2 \in Y, x_1 \neq x_2 \\
1 & \text{otherwise,}
\end{cases}$$
for $i = 1, 2$. To show these are both metrics, the only possible trap here is triangle inequality, as the new points could "short-cut" the old points. However, this is not a problem, as $d_1, d_2$ are bounded above by $2$, i.e. there is no $m, n \in \Bbb{N}_0$ such that $d_i(m, n) > 2$, for $i = 1, 2$.
If $\operatorname{id} : (X, d'_1) \to (X, d'_2)$ were continuous, then the restriction to $Y$ would be continuous, which, given our definition of $d'_1$ and $d'_2$, would make the identity on $\Bbb{N}_0$ continuous between $d_1$ and $d_2$. This is untrue, and so we are done.
