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Let $\mathbb{S}$ be the Sierpinski space, that is $\mathbb{S}$ has $\{0,1\}$ as a base set, and $\tau = \{\emptyset, \{0\}, \{0,1\}\}$ as a topology.

The Sierpinski space $\mathbb{S}$ has the following property:

$(\star)$ Given any topological spaces $X,Y$ and a function $f:X\to Y$, then $f$ is continuous if for every continuous map $z:Y\to \mathbb{S}$ the map $z\circ f: X\to\mathbb{S}$ is continuous.

(Proof. If $f: X\to Y$ is not continuous, then there is $V\subseteq Y$ open such that $f^{-1}(V)$ is not open in $X$. Defining $z_V: Y\to \mathbb{S}$ as mapping $V$ to $0$ and $Y\setminus V$ to $1$ we immediately see that $z_V\circ f: X \to Z$ is not continuous.)

If a topological space $S$ has property $(\star)$ then we call it Sierpinski-like. (Maybe there is some standard terminology, but I wasn't able to find it.)

Soft question: How can Sierpinski-like spaces be characterized?

Concrete question: Do Sierpinski-like spaces have to be $T_0$? Or do they necessarily contain a non-empty open set $U_0$ such that $U_0\subseteq U$ for all open non-empty $U$?

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    $\begingroup$ Any space containing $\mathbb S$ as a subspace will work, hence the answer to both “concrete questions” is negative. $\endgroup$ – Emil Jeřábek Jan 6 '15 at 15:18
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    $\begingroup$ On second thought, that’s an if and only if characterization, actually. $\endgroup$ – Emil Jeřábek Jan 6 '15 at 15:27
  • $\begingroup$ Hmm I see - thanks! Maybe I should close the question - or alternatively, you can put your thoughts as an answer. $\endgroup$ – Dominic van der Zypen Jan 6 '15 at 15:29
  • $\begingroup$ I always knew your Sierpiński's space as P.S.Alexandrov's space. Do you know the (hi)story? $\endgroup$ – Włodzimierz Holsztyński Jan 6 '15 at 16:40
  • $\begingroup$ The Sierpiński space is a prototypical Alexandrov space in the sense topospaces.subwiki.org/wiki/Alexandrov_space . Maybe that’s the source of the variant terminology? $\endgroup$ – Emil Jeřábek Jan 6 '15 at 17:19
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The following are equivalent:

  1. $S$ is Sierpiński-like.

  2. $\mathbb S$ embeds in $S$ as a subspace.

  3. $S$ is not symmetric (i.e., $R_0$).

$2\to1$ follows from the facts that $\mathbb S$ is Sierpiński-like, and if $Y\subseteq Z$ is a subspace and $f\colon X\to Y$, then $f$ is continuous as a mapping $X\to Y$ iff it is continuous as a mapping $X\to Z$.

$2\leftrightarrow3$ is a restatement of the definition of $R_0$.

$1\to3$: If $S$ is $R_0$, then the points in the range of any continuous mapping $z\colon\mathbb S\to S$ are topologically indistinguishable, hence $z\circ f$ is continuous for any mapping $f\colon X\to\mathbb S$. Thus, taking any discontinuous mapping $X\to\mathbb S$ shows that $S$ is not Sierpiński-like.

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    $\begingroup$ I am not familiar with the non-symmetric*/$R_0 notion. Would you add it to your Answer? $\endgroup$ – Włodzimierz Holsztyński Jan 6 '15 at 16:38
  • $\begingroup$ This is not a terribly common separation axiom, it means that the Kolmogorov quotient is $T_1$. I’ve added a link to the definition. $\endgroup$ – Emil Jeřábek Jan 6 '15 at 17:05
  • $\begingroup$ Emil, thank you (there is no end to the topological separation axioms :-) $\endgroup$ – Włodzimierz Holsztyński Jan 6 '15 at 17:33

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