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I am trying to follow Theorem (3.11) of Kechris's Classical Descriptive Set Theory. In this part of the proof he shows that a $G_{\delta}$-subspace Y of a completely metrizable space $(X,d)$ is completely metrizable. For this, he defines the following metric. Sadly, he skips over why this new metric induces the subspace topology. Here is the relevant part of the proof: (screenshot)

$\quad$ For the second assertion, let $Y=\bigcap_{n} U_{n}$, with $U_{n}$ open in $X$. Let $F_{n}=X \backslash U_{n}$. Let $d$ be a complete compatible metric for $X$. Define a new metric on $Y$, by letting $$ d^{\prime}(x, y)=d(x, y)+\sum_{n=0}^{\infty} \min \left\{2^{-n-1},\left|\frac{1}{d\left(x, F_{n}\right)}-\frac{1}{d\left(y, F_{n}\right)}\right|\right\} . $$ It is easy to check that this is a metric compatible with the topology of $Y$.

Because of $d'(x,y)\geq d(x,y)$ it follows that any for $\epsilon > 0$ and any $x\in X$, $B_{d'}(x, \epsilon) \subset B_d(x, \epsilon)$. So any set open in $(X, d)$ is also open in $(Y, d')$. But I can't figure out the other direction. Any help is appreciated.

Kechris, Alexander S., Classical descriptive set theory, Graduate Texts in Mathematics. 156. Berlin: Springer-Verlag. xx, 402 p. (1995). ZBL0819.04002.

Calvin Khor
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a_hayler
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    For a different (hopefully easier) argument you can consider the map $Y\to X\times\Bbb R^\Bbb N$ given by $y\mapsto (y, (1/d(x,F_n)_n)$ and check that this is a continuous embedding with closed image (closed subspaces of Polish spaces are clearly Polish themselves) – Alessandro Codenotti Jan 16 '22 at 15:43
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    @AlessandroCodenotti That embedding is actually where this metric comes from. It's the product metric (series variant) restricted to that subset. – Henno Brandsma Jan 17 '22 at 09:02

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Let's abstract away from the precise formula for now. We have uniformly continuous functions (uniformly because for fixed $n$ $d(x,F_n)$ is uniformly continuous and bounded away from $0$ so the inverse is too) $f_n: Y \to [0,\frac{1}{2^{n+1}}]$ so that

$$d'(x,y)= d(x,y) + \sum_{n=0}^\infty |f_n(x) - f_n(y)|$$

Let $r>0$ be arbitrary. Then we can find $\delta>0$ (wlog $\delta < r$) so that for $x,y \in Y$, $d(x,y) < \delta$ implies $\sum_{n=0}^\infty |f_n(x) - f_n(y)| < r$, say. This uses the convergence of the series and the uniform continuity of the $f_n$.

So $$\forall r>0 :\exists \delta>0: \forall x,y \in Y: d(x,y) < \delta \implies d'(x,y) < 2r$$

Show that this implies the other inclusion of metric topologies.

Henno Brandsma
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  • Could you explain why $d(x, F_n)$ has to be bounded away from $0$? For example, with $U_n$ being $(0,1)$ for all $n$. Then I would say that any $\epsilon > 0$ is in $Y$ and I can get a distance as small as I want. – a_hayler Jan 21 '22 at 11:10
  • @a_hayler if the distance can get arbitrarily close to $0$, one over the distance explodes and we lose uniform continuity. $\frac1x$ is uc on an interval $[a, 4)), a>0$ not on $(0,4]$ e.g. – Henno Brandsma Jan 21 '22 at 11:17
  • @Henno_Brandsma I maybe did not formulate my question correctly. Why can I assume in the proof that the question is referring to that $d(F_n,x)$ is bounded? – a_hayler Jan 21 '22 at 11:23
  • @a_hayler you’re not assuming it’s bounded at all. The reciprocal is and that’s provable. – Henno Brandsma Jan 24 '22 at 17:03