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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.

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  • $\begingroup$ What you've written is extremely sloppy, so I'm going to critique it. I hope you don't mind - the point is to help you write more clearly in the future. First line after My logic: There is no fixed $f$. The point is that for any $f$, $f^{-1}(U)$ is open. Second line: This is not a complete sentence. Third line: 1. $W\in (X,T)$ isn't what you meant. You probably meant $W\in T$ or $W\subseteq X$ is open. 2. What are the domain and codomain of $g$? You've written $g^{-1}(W) = V$, so $g$ must be a function $X\to X$, but then you're not using $(\mathbb{R},st)$ at all. $\endgroup$ – Alex Kruckman Mar 6 '18 at 21:03
  • $\begingroup$ Fourth line: "We can do this because of continuity" - continuity of what? "We can just form a union or intersection of $W$s which generated this $V$ through the preimage of $g$". I can't make any sense of this at all. $\endgroup$ – Alex Kruckman Mar 6 '18 at 21:06
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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 $f_E^{-1}((0,\infty))$?

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  • $\begingroup$ Is it just $E$? $\endgroup$ – Gregory Peck Mar 6 '18 at 20:59
  • $\begingroup$ Yes, exactly. So since $f_E$ is continuous and $(0,\infty)$ is open in $\mathbb{R}$, $E$ must be open. $\endgroup$ – carmichael561 Mar 6 '18 at 20:59

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