# $(X, \mathcal{T})$ Polish and $F \subseteq X$ closed $\implies$ topology generated by $\mathcal{T} \cup \{ F \}$ is Polish

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?

• 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. Jun 28, 2021 at 15:02
• @Sven-OleBehrend Thank you. But isn't this a contradiction to this statement? math.stackexchange.com/questions/1526343/… Jun 28, 2021 at 15:21
• @qwertz That linked question was for non-disjoint union.
– Alan
Jun 28, 2021 at 15:58

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?