Let all $f:(X,T) \rightarrow (\mathbb{R}, st)$ be continuous. Show $T$ discrete.

Question

Let all $f:(X,T) \rightarrow (\mathbb{R}, st)$ be continuous for every $f$. Show $T$ discrete.

$st$ is the standard topology.

My logic:

So we know that for all $U \in (\mathbb{R}, st)$, that $f^{-1}(U)$ is open in $(X,T)$

Given any $V \subset X$.

Then there exists a continuous function $g$ such that $g^{-1}(W) = V$ for some $W \in (X,T)$

We can do this because of continuity, we can just form a union or intersection of $W$s which generated this $V$ through the preimage of $g$.

Very wishy washy I know. But is my intuition correct? Can we make it more rigorous?

It was trivial for me to show That given $T$ discrete that all functions are continuous. The other way around isn't so trivial.

Answer 1

I think you're on the right track. To make the argument more precise, for each subset $E$ of $X$ consider the function $f_E$ defined by $f_E(x)=1$ if $x\in E$ and $f_E(x)=0$ otherwise. What is $f_E^{-1}((0,\infty))$?