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Let $X$ be a set with the property that for any two metrics $d_1$, and $d_2$ on $X$, the identity map $\operatorname{id} : (X, d_1) \to (X, d_2)$ is continuous. Which of the following are true?

(a) $X$ must be a singleton.

(b) $X$ can be any finite set.

(c) $X$ cannot be infinite.

(d) $X$ may be infinite but not uncountable.

My attempt: If the identity map is continuous then topology generated by $d_2$ is contained in the topology generated by $d_1$. Similarly, topology generated by $d_1$ is contained in the topology generated by $d_2$. Take $d_1$ as discrete metric and $d_2$ as indiscrete metric. The two topologies must be the same which implies $X$ must be singleton.

This question has already been asked but I am not convinced with the answer:

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  • $\begingroup$ The question you linked to seems to have misstated the problem. I suggest you post a comment about it. $\endgroup$ – Rob Arthan Apr 8 '15 at 20:21
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for a finite set any metric space is basically discrete metric space... so it is true.

let X be an infinite set...so there is an injection from natural numbers to X...let Y={$y_n$} be a countably infinite subset of X... then $X= Y\cup Y^c$ ...define a metric $d$ on $X$ as follows... $d(y_n,y_0)=1/n$ , $d(y_n,y_m)= |1/n - 1/m|$ , $d(y_i,x)=1$ ,$d(y,z)=1$ where $x,y,z \in Y^c$...then $d$ is a metric on $X$..consider $d_1$ as the discrete metric space...then cnsider $id: (X,d) \to (X,d_1)$ then it is not continuous...since $ y_n \to y_0$ as $n \to \infty$ in $d$ but $y_n$ doesnot converge to $y$ in $d_1$. so it is not true if $X$ is infinite.

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Your question appears to differ from the linked one in that you require the statement to be true for all possible pairs of metrics $d_1, d_2$.

Note that metric topologies are always Hausdorff, so the indiscrete topology is not metrizable and in particular if $X$ is finite the statement is true.

If $X$ has any metric other than the discrete metric (you should check that this is true for any infinite set), then set that metric as $d_1$ and the discrete metric as $d_2$ for your counterexample. (Note that all maps with domain a discrete space or codomain an indiscrete space are continuous, so your proposed counterexample needs $d_1$ and $d_2$ to be switched.)

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The identity mapping on a set $X$ is a continuous mapping with respect to any two metrics on $X$ iff $X$ is finite. Hence (a) and (d) are false while (b) and (c) are true.

If $X$ is finite, then as metric topologies are Hausdorff and there is only one Hausdorff topology on $X$, namely the discrete topology, any two metrics on $X$ induce the same topology and the identity mapping is continuous with respect to that topology.

If $X$ is infinite, you can choose a countably infinite subset $Y = \{y_0, y_1, y_2, \ldots\}$ of $X$, and a metric $d_1$ on $X$ such that $y_0$ is an accumulation point of $Y$ and all other points of $X$ are isolated (to do this, identify $Y$ with the subset $\{0\} \cup \{\frac{1}{n} \mathrel{|} 0 < n \in \mathbb{N}\}$ of $\mathbb{R}$ to define the metric on $Y$ and define $d(x, y) = 1$ whenever $x \not= y$ and $\{y, x\} \not\subseteq Y$). If $d_2$ is the discrete metric on $X$, the identity mapping cannot be a continuous mapping from $(X, d_1)$ to $(X, d_2)$.

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