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Construct an example of a topological space $(X,T)$ that is not sequential and is not $T_0$.

Preferably the example should not involve a pseudometric, a finite set $X$, or the trivial topology $\{X, \emptyset\}$

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    $\begingroup$ Disjoint union of a non-sequential space with a space that's not $T_0$? $\endgroup$ Sep 4, 2015 at 23:49
  • $\begingroup$ @silvascientist: Or the product. $\endgroup$ Sep 5, 2015 at 19:35
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    $\begingroup$ So whose idea was it to undelete this post? $\endgroup$
    – cpiegore
    Jul 19, 2016 at 1:26
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    $\begingroup$ I do not know how to interpret your earlier comment. Do you prefer for your post to be deleted or undeleted? $\endgroup$
    – quid
    Aug 31, 2016 at 23:42
  • $\begingroup$ @quid Tell me something: do you think there is anything good about this post or is it all bad, ignoring the fact that it was closed and deleted? I understand why it was closed and deleted the first time, but I don't get why it was undeleted and "re-deleted" several times. $\endgroup$
    – cpiegore
    Sep 1, 2016 at 1:37

1 Answer 1

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Let $Y=\omega_1+1$ with the order topology $\tau'$, and let $p$ be a point not in $Y$. Let $X=Y\cup\{p\}$, and let

$$\tau=\{U\in\tau':0\notin U\}\cup\big\{U\cup\{p\}:0\in U\in\tau'\big\}\;$$

then $\langle X,\tau\rangle$ is not $T_0$, because every open set contains either both or neither of the points $0$ and $p$, and it’s not sequential, because $X\setminus\{\omega_1\}$ is a sequentially closed set that is not closed.

Added: In case you’re not familiar with ordinals and their topology, here’s a simpler variant of the same basic idea. Let $Y$ be an uncountable set, $y_0$ and $y_1$ distinct points of $y$, and $p$ a point not in $Y$. Let $X=Y\cup\{p\}$, $\mathscr{U}=\wp(Y\setminus\{y_0,y_1\})$, and $\mathscr{V}=\big\{U\cup\{y_0,y_1\}:U\in\mathscr{U}\big\}$. Finally, let

$$\tau=\mathscr{U}\cup\mathscr{V}\cup\{X\setminus C:C\in\mathscr{U}\cup\mathscr{V}\text{ and }C\text{ is countable}\}\;;$$

then $\tau$ is a topology on $X$. Every member of $\tau$ contains either both or neither of the points $y_0$ and $y_1$, so $\langle X,\tau\rangle$ is not $T_0$. $Y$ is sequentially closed, since the convergent sequences in $Y$ are those that are eventually constant or eventually in the set $\{y_0,y_1\}$, but $p\in\operatorname{cl}_XY$, so $Y$ is not closed in $X$.

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    $\begingroup$ If I understand the solution correctly, isn't the basic idea simply taking any space $Y$ which is not sequential and then removing $T_0$-property by "doubling" one point. (I.e., adding a new point $p$ and creating neighborhood of this point from some $y\in Y$ simply by using both these points in all open sets of the original space containing $y$)? $\endgroup$ Sep 1, 2016 at 2:47
  • $\begingroup$ @Martin: Yes. I don’t remember whether I consciously realized that at the time and wanted to stick with concrete examples, or whether I didn’t consciously notice the generalization. I’m pretty sure that I doubled one of the isolated points in the second example (rather than the limit point) because that made it extremely easy to see why $Y$ is sequentially closed. $\endgroup$ Sep 1, 2016 at 3:11

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