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Let $X$ be uncountable set with the co-countable topology. How do continious maps $X \rightarrow X $ look like?

So by the definition what is a co-countable topology?

Co-countable topology on set $X$ is defined as follows:

$T_c$ is a collection of all subsets $U$ of $X$ such that $X \setminus U$ is either countable or it is all of $X$.

And continuous map $f:X \rightarrow Y$ is a map such that if we take any $U \subset Y$ such that $U$ is open in Y $\Rightarrow$ $f^{-1}(U)$ is open in $X$

So if we take any basis element $U$ of $X$ it is such that either $X\setminus U$ is countable or $X$ or an empty set. And it has to be mapped to an element such that $X\setminus U$ is countable or $X$ or an empty set as well.

So can anyone help please how to proceed next? How do this functions look like?

Definitely, identity map is continuous. But is there any other continuous functions between $X$ and $X$?

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  • $\begingroup$ * continuous $\endgroup$ – Arnaud Mortier Apr 30 '18 at 19:04
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A functions $f\colon X\longrightarrow X$ is continuous if and only if, for every closed subset $C$ of $X$, $f^{-1}(C)$ is also closed. This means that $f^{-1}(X)=X$ (this is always true) and that if $C$ is a countable set, then $f^{-1}(C)$ is also a countable set. Therefore, if $f$ is not constant, then $f$ is continuous if and only if there is no infinite uncountable set $A\subset X$ such that $f(A)$ is countable.

So, for instance, if $X=\mathbb R$ , the function $f(x)=\max\{|x|,1\}$ is not countinuous, because $f([-1,1])=\{1\}$.

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    $\begingroup$ It's enough that for every $x \in X$, $f^{-1}(\{x\})$ is countable. $\endgroup$ – Alex Zorn Apr 30 '18 at 19:11
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    $\begingroup$ @AlexZorn Yes, you're right. I should have thought that. $\endgroup$ – José Carlos Santos Apr 30 '18 at 19:47
  • $\begingroup$ f([-1,1]) = [0,1]. $\endgroup$ – William Elliot Apr 30 '18 at 21:20
  • $\begingroup$ @WilliamElliot I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Apr 30 '18 at 21:22

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