Let $ ( X , \tau )$ be a $T_2$ topological space such that every open continuous image of $X$ is $T_2$. Show that $ \tau$ is the discrete topology on $X$.

This is a question I have been thinking about and feel I am getting nowhere. The only thing I could think to try was using the theorem which says:

Lemma. If $f$ is a continuous open map of $X$ onto $Y$, then $Y$ is Hausdorff iff $ \{ (x_1, x_2) : f(x_1) = f(x_2) \}$ is a closed subset of $X \times X$.

I tried looking for some map that might let me use the lemma above to prove that any given set $A \in \tau$ is such that $ A \times A$ is closed by the lemma, and therefore $A$ is closed . I haven't had any success. Any hint or help would be appreciated, thanks!

  • $\begingroup$ What do you mean by "open continuous image"? $\endgroup$
    – Rob Arthan
    Aug 5, 2015 at 23:35
  • $\begingroup$ @RobArthan By that I mean that it is the image of continuous map which is also open ( i.e. for any open set in the original space $X$ say $U$, $f(U)$ is open in the image). $\endgroup$
    – Jmaff
    Aug 5, 2015 at 23:49
  • $\begingroup$ I think you should clarify that in your question and title: "image under a continuous open map" rather than "open continuous image". I guessed that was what you meant, but it isn't obvious. $\endgroup$
    – Rob Arthan
    Aug 5, 2015 at 23:56
  • $\begingroup$ sounds good, thanks. $\endgroup$
    – Jmaff
    Aug 5, 2015 at 23:59

1 Answer 1


Let $\mathbb S$ be the Sierpiński space, i.e. $\mathbb S=\{0,1\}$ equipped with the topology $\{\emptyset,\{1\},\mathbb S\}$. For $a\in X$, define a map $f_a:X\to\mathbb S$ by $$f_a(x)=\begin{cases}0;&x=a,\\1&x\neq a.\end{cases}$$ The map $f_a$ is continuous, because $f^{-1}\{1\}=X\setminus\{a\}$ must be open, since points are closed in Hausdorff spaces.

Also note that if $X=\{a\}$, then $X$ is clearly discrete, so we shall assume that $X$ has at least $2$ points, in which case $f_a$ is surjective.

The map $f_a$ is open if and only if $a$ is not open. To see this, notice that $\{0\}$ is the only non-open subset of $\mathbb S$, so $f_a$ can only fail to be open if $\{0\}$ is the image of some open set. Now, $\{a\}$ is the only set $f_a$ takes to $\{0\}$, so $f_a$ can only fail to be open if $\{a\}$ is open.

We shall now prove that every singleton $\{a\}\subseteq X$ is open, thus establishing the claim that $X$ is discrete. We argue by contradiction: suppose $\{a\}$ is not open. Then $f_a$ is open and continuous. Therefore, its image $\mathbb S$ is a Hausdorff space, a clear contradiction.

This contradiction proves that points in $X$ are open, so $X$ is discrete.


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