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).


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.)

  • $\begingroup$ 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
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    $\begingroup$ 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

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