What would be examples of topological spaces that are neither sequential nor $T_0$?

$\pi$-Base does not have an example of such space.

I have seen some examples of non-sequential spaces. Arens-Fort space is mentioned in this post. Another space mentioned there is the Stone-Čech compactification $\beta\mathbb N$. However, both of them are Hausdorff.

From non-Hausdorff spaces I was able to find the cocountable topology on an uncountable set. However, it is still a $T_1$-space.

So, the question remains of giving simple examples of non-sequential spaces which are not $T_0$.

  • 2
    $\begingroup$ The reason for posting this was the delete-undelete war on this post. (See also meta.) This also means that I know the answer - I have seen it on the deleted post. The repeated deletion and undeletion suggests that some users consider this question useful and some do not. I hope that this version of the question has the better form than the original one. Perhaps in this way everybody is satisfied. $\endgroup$ Sep 2, 2016 at 21:32
  • $\begingroup$ Can't you just pick a point $x_0$ that is in your favourite non-sequential space $X$ and a point $x_1$ that is not and then topologise the disjoint union $Y = X \sqcup \{x_1\}$, by taking $A\subseteq Y$ to be open if $A \cap X$ is open in $X$ and either (1) $x_0 \not\in A$ or (2) $\{x_0, x_1\} \subseteq A$? Then every open subset of $Y$ that contains $x_0$ also contains $x_1$, so $Y$ is not $T_0$, while a sequentially open subset of $X$ that is not open will still by sequentially open but not open in $Y$. $\endgroup$
    – Rob Arthan
    Sep 2, 2016 at 22:10
  • $\begingroup$ @RobArthan Yes, that seems like a rather natural way to construct such an example. In fact, this is precisely the space described in Brian M. Scott's answer in the paragraph about "a very simple machine". $\endgroup$ Sep 2, 2016 at 22:23
  • $\begingroup$ Indeed: I was typing my comment while Brian was typing his answer, $\endgroup$
    – Rob Arthan
    Sep 2, 2016 at 22:28
  • $\begingroup$ @MartinSleziak Well played. $\endgroup$
    – user99914
    Sep 3, 2016 at 2:41

1 Answer 1


For a very elementary example 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 $X$.

One that fails to be sequential but still has non-trivial convergent sequences can be obtained by letting $Y=\omega_1+1$ with the order topology $\tau'$ and $p$ 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.

Each of these is the result of applying a very simple ‘machine’ to an arbitrary non-sequential space $\langle Y,\tau\rangle$. Fix some $y_0\in Y$, let $Z=\{0,1\}$ have the indiscete topology, and let $X=(Y\times Z)/{\sim}$, where $\langle y_1,z_0\rangle\sim\langle y_2,z_1\rangle$ if and only if either $\langle y_1,z_0\rangle=\langle y_2,z_1\rangle$, or $y_1=y_2\ne y_0$. This simply doubles the point $y_0$, turning it into two distinct points with identical nbhds in $X$, in such a way that $Y$ is the Kolmogorov quotient of $X$.

In fact it is true in general that if $X$ is non-sequential, then the Kolmogorov quotient $Y$ of $X$ is non-sequential, so every example of a non-sequential, non-$T_0$ space can be obtained by ‘fattening up’ a $T_0$ non-sequential space, albeit not necessarily at just one point.


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