6
$\begingroup$

Schwartz writes in his book "Radon Measures on Arbitrary Topological Spaces", p. 110: "But even a finite union of polish spaces need not be polish". The same statement can be also found here.

How should this statement be interpreted? If I have a Polish space $X$ and a finite union of Polish subspaces $X_i \subseteq X$ then all of them are $G_\delta$-subsets and their (finite) union is again a $G_\delta$-subset of $X$ thus a Polish space.

Should this statement be interpreted that the given Polish spaces $X_i$ are not necessarily contained in a joint Polish space $X$? But what topology do we impose on the union? It may not be a disjoint union (which would be again Polish with the disjoint union topology).

$\endgroup$
2
$\begingroup$

Example of a non-metrizable space $S$ with two subspaces $S_1, S_2 $, each homeomorphic to the real line $R$, such that $S=S_1\cup S_2$ .$$\text {Let } S= (Q\times \{0\})\cup ((R\backslash Q)\times \{1,2\})$$ where $Q$ is the rationals. Let $T$ be the usual topology on $R$. $$\text {For } t\in T \text { let } t^*=((t\cap Q)\times \{0\})\cup ((t\backslash Q)\times \{1,2\})=(t\times \{0,1,2\})\cap S.$$ $$\text {Let } B=\{t^*\backslash u :t\in T\land ( u \text { is finite})\}.$$ I will leave it to you verify the following :(1) $B$ is a base for a topology $V$ on $S$. (2) $V$ is not a Hausdorff topology.(3)With the topology $V$ on $S$,the subspaces $$S_j=(Q\times \{0\})\cup ((R\backslash Q)\times \{j\}), \text { for } j\in \{1,2\}$$ are each homeomorphic to $R$ (by projection onto the first co-ordinate.)

$\endgroup$
  • $\begingroup$ I see, a finite union of Polish subspaces of a non-Polish space might be not Hausdorff. Nice example! Btw, a simpler example in the same spirit as your example can be given by considering e.g. two "parallel" real lines $\mathbb{R} \times \{ 1, 2 \}$ with open sets $t \times \{ 1, 2 \}$, $t$ open in $\mathbb{R}$ or the line with two origins $0_1$ and $0_2$. Both have $\mathbb{R}$ as their Kolmogorov quotient. $\endgroup$ – yadaddy Nov 13 '15 at 6:44
  • 1
    $\begingroup$ I found this example once in a different context: If $X$ is completely metrizable (c.m.) and $Y\subset X$ then $Y$ is c.m. iff $Y$ is $G_{\delta}$ in $X$. One step in proving it is that if $F$ is a countable set of c.m. subspaces of $X$ then $\cap F$ is c.m. I saw that this step required only that $X$ is $T_2$. The space $S$ is $T_1$,not $T_2$ ; the subspaces $S_1, S_2$ are c.m, but $S_1\cap S_2$ which is homeomorphic to $Q$,is not c.m. $\endgroup$ – DanielWainfleet Nov 13 '15 at 6:54
  • $\begingroup$ So this example then also shows that the finite intersection of Polish subspaces of a topological space need not be Polish (even when assuming that the intersection is not empty). $\endgroup$ – yadaddy Nov 13 '15 at 7:04

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.