I am trying to understand the proof of this statement (Kechris, Classical Descriptive Set Theory, p. 82).

(13.2) Lemma. Let $(X, \mathcal{T})$ be Polish and $F \subseteq X$ closed. Let $\mathcal{T}_F$ be the topology generated by $\mathcal{T} \cup \{ F \}$. Then $\mathcal{T}_F$ is Polish [...].

Proof. Note that $\mathcal{T}_F$ is the direct sum of the relative topologies on $F$ and ${\sim}F$ so, by 3.11, $\mathcal{T}_F$ ist Polish.

With theorem 3.11 stating the following.

(3.11) Theorem. If $X$ is metrizable and $Y \subseteq X$ is completely metrizable, then $Y$ is a $G_\delta$. Conversely, if $X$ is completely metrizable and $Y \subseteq X$ is a $G_\delta$, then $Y$ is completely metrizable. In particular, a subspace of a Polish space is Polish iff it is a $G_\delta$.

The proof is a rather blurry. I tried to connect the dots and made the following observations:

  1. $\mathcal{T} \cup \{ F \}$ is a subbasis of $\mathcal{T}_F$, thus $\mathcal{T} \cup \mathcal{T}|F$ is a subbasis for $\mathcal{T}_F$, thus $\mathcal{T}|F \cup \mathcal{T}|{\sim}F$ is a basis for $\mathcal{T}_F$.

  2. It follows from 1. that $U \in \mathcal{T}_F$ iff $U \cap F \in \mathcal{T}|F$ and $U \cap {\sim}F \in \mathcal{T}|{\sim}F$.

  3. Since 2. and $F$ and ${\sim}F$ are disjoint, $\mathcal{T}_F$ is the direct sum (disjoint union) of the relative topologies on $F$ and ${\sim}F$.

  4. Theorem 3.11 implies: Since $F$ and ${\sim}F$ are $G_\delta$ in the Polish space $(X, \mathcal{T})$, the spaces $(F, \mathcal{T}|F)$ and $({\sim}F, \mathcal{T}|{\sim}F)$ are Polish.

  5. It follows from 2. that $\mathcal{T}|F = \mathcal{T}_F|F$ and $\mathcal{T}|{\sim}F = \mathcal{T}_F|{\sim}F$, thus $(F, \mathcal{T}_F|F)$ and $({\sim}F, \mathcal{T}_F|{\sim}F)$ are Polish with 4.

Where do I make the step to $\mathcal{T}_F$ being Polish? As far as I can see, it does not follow from 3.11. Do I need another theorem?

  • $\begingroup$ You can (and probably have shown in this book) that the finite (or even countable) disjoint union / sum of Polish spaces is again polish. Therefore $\mathcal T_F \cong T|F \oplus T|∼F$ is polish. $\endgroup$ Jun 28, 2021 at 15:02
  • $\begingroup$ @Sven-OleBehrend Thank you. But isn't this a contradiction to this statement? math.stackexchange.com/questions/1526343/… $\endgroup$
    – qwertz
    Jun 28, 2021 at 15:21
  • $\begingroup$ @qwertz That linked question was for non-disjoint union. $\endgroup$
    – Alan
    Jun 28, 2021 at 15:58

1 Answer 1


Look at the two sets $F$ and $X\setminus F$ (or $\sim F$ if you prefer), in their subspace topology induced from $\mathcal{T}$.

If we introduce a new open set $F$ (and generate the topology from $\mathcal{T} \cup \{F\}$, called $\mathcal{T}_F$) then on $F$ itself nothing really changes in its relative topology wrt $\mathcal{T}_F$ compared to $\mathcal{T}$ ($F$ is already open in the relative topology, of course). And on $X\setminus F$ we also have that $\mathcal{T}_F$ restricted to $X\setminus F$ is the same as $\mathcal{T}$ restricted to that same subspace.

So mapping $x \in X$ to the same point in $F \oplus (X\setminus F)$ is a continuous open bijection, almost by definition.

So as $F$ is closed It's subspace topology is Polish and the same holds for the open subspace $X\setminus F$ by your quoted theorem (open is $G_\delta$, as are closed sets) and it's easy to see that the sum space of two Polish spaces is Polish (just definitions: completeness and separability and metrisability are clear).

So you're only using 3.11 for the open subspace part, I suppose, though the text seems to suggest it's used for the sum part?


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