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Let $X$ denote a countable infinite set endowed with complement topology (that is, $X$ itself and all its finite subsets consisit of the set of closed set). Can we find a Hausdorff compact space $Y$ such that there is a subjective continuous map from $Y$ to $X$?

I come across this question when trying to decide if $X$ is compactly generated. Any idea will be appreciated.

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Suppose $X$ is an infinite set and let $\mathcal{T}_{cf} = \{ U\subset X\ |\ X\setminus U\ \mathrm{is\ finite\ or}\ U=\emptyset\}$. Let $p$ be an element that is not in $X$ and let $Y=X\cup \{p\}$.

Topologize $Y$ by declaring a set $U$ in $Y$ to be open if and only if $U\subset X$ or if $p \in U$, then $X\setminus U$ must be finite. $Y$ is the Alexandrov compactifaction or also sometimes called the one-point compactification of the discrete space $(X,\mathcal{P}(X))$. I leave it to you to verify that $Y$ is a compact Hausdorff space.

Let $x_0$ be a fixed member of $X$ and define $f:Y\rightarrow X$ as follows: $$f(a) = \begin{cases} x & \mathrm{if}\ a=x\in X \\ x_0 & \mathrm{if} \ a=p. \end{cases}$$

$f$ is the desired continuous surjection. Let $V$ be an open set of $Y$, then: $$f^{-1}(V) = \begin{cases} V & \mathrm{if}\ x_0\not\in V \\ V\cup\{p\} & \mathrm{if} \ x_0 \in V. \end{cases}$$

If $x_0 \not\in V$, then $f^{-1}(V)$ is trivially open in $Y$, so we need only check the other case. Without loss of generality, we may assume that $\emptyset \subsetneq V\subsetneq X$, then $X\setminus V$ must be a non-empty finite set, say $X\setminus V= \{x_1, \ldots , x_n\}$.

Consequently $X\setminus f^{-1}(V) = X\setminus (V\cup \{p\})= (X\setminus V) \cap X\setminus \{p\}= X\setminus V$ is finite and hence closed, thus $f$ is continuous.

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