1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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

0
$\begingroup$

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?

$\endgroup$

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.