Let $(X,\tau)$ be a topological space. Do the set of all $X\rightarrow X$ continuous functions uniquely determines $\tau$?

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    $\begingroup$ No. But the set of all continuous mapping $X\to\Bbb S$ does, where $\Bbb S$ is the Sierpiński space. See here. $\endgroup$
    – Dejan Govc
    Jun 13 '12 at 13:12

No. For example, the discrete and trivial topologies both make every function from $X$ to $X$ continuous.



let $X$ be an arbitrary set with two or more elements, and let $\tau_1$ be the discrete topology on $X$ (all sets are open), and let $\tau_2$ be the trivial topology on $X$ (only $X$ and $\emptyset$ are open). Note that all maps from $(X,\tau_1)$ and all maps into $(X,\tau_2)$ are continuous, regardless of where they go or where they come from respectively. Thus, in both cases, the set of all continuous maps $X\rightarrow X $ is the set of all set-theoretic maps $X\rightarrow X $, but $\tau_1 \neq \tau_2$.


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