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 $U$?

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.