Sachchidanand Prasad

09-02-2023

Problem: Let $(X,d)$ be any metric space. Suppose that every function \[ f: (X,d) \to \left( \mathbb{R}, d_{\textup{Euc}} \right) \] is continuous, where $d_{\textup{Euc}}$ is the Euclidean metric on $\mathbb{R} $. Then prove that $d$ is the discrete metric.

Solution: If $X$ has only one point, then it has has the discrete metric. So we assume that $X$ has at least two points. Let $p\in X$. Consider the function \[ f_p:X\to \mathbb{R} ,~ \begin{dcases} 0, &\text{ if } p\in X ;\\ 1, &\text{ if } p\notin X. \end{dcases} \]

According the given hypothesis, the function $f_p$ is continuous for every $p\in X$. So, we have \[ f^{-1}(\{0\}) =\{p\},~\text{and }~ f^{-1}(\{1\}) = X\setminus \{p\}. \] Since $f_p$ is continuous, and $\{1\}$ is a closed set, so $f^{-1}(\{1\}) = X\setminus \{p\}$ will be closed and hence $\{p\}$ will be open. So we showed for every $p\in X$ the set $\{p\}$ is open. We have the following result which proves that $d$ is the discrete metric.

Let $(X,d)$ be a metric space such that every singleton is open. then $d$ is the discrete metric.
Since every singleton set is open and we can write any set as unions of singleton, so any set is open. More precisely, let \[ U = \bigcup_{\alpha \in A} \left\{x_\alpha\right\}. \] Since each $\left\{x_\alpha\right\}$ is open and arbitrary union of open sets is open, so $U$ is open. Thus every set is open and hence $d$ is the discrete metric.

The above problem is true even $(X,\mathcal{T})$ is a topological space. The solution will be same.