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Let $X$ be a non-empty topological space such that every function $f:X\to\mathbb R$ is continuous , then is every subset of $X$ open ?

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Pick any $x\in X$ and define $f:X\to\mathbb R$ as follows: \begin{align*} f(y)=\begin{cases}1&\text{if $y=x$,}\\0&\text{if $y\in X$, $y\neq x$.}\end{cases} \end{align*} Then, the interval $(1/2,3/2)$ is open in $\mathbb R$, so $\{x\}=f^{-1}((1/2,3/2))$ is open in $X$, since $f$ is continuous by assumption. Therefore, every singleton in $X$ is open, so the topology in question must be the discrete one.

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Given $A\subseteq X$, consider the mapping $f:X\to \Bbb R$ that sends every element of $A$ to $1$ and every element of its complement to $0$. Then?

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  • $\begingroup$ Might this proof even work for any set we map to with $\geq 2$ elements? $\endgroup$ – GPerez Dec 29 '14 at 15:58
  • $\begingroup$ @GPerez It suffices there is some space with more than two points for which every mapping is continuous. $\endgroup$ – Pedro Tamaroff Dec 29 '14 at 15:59
  • $\begingroup$ It might actually have to be $T_0$, I think. $\endgroup$ – GPerez Dec 29 '14 at 16:10
  • $\begingroup$ For instance, $\{a,b\}$ and the coarse topology won't prove anything about $X$. $\endgroup$ – GPerez Dec 29 '14 at 16:12
  • $\begingroup$ @GPerez You're correct. It suffices singleton sets are closed, for example. $\endgroup$ – Pedro Tamaroff Dec 29 '14 at 16:36

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