In Alexander Kechris' Classical Descriptive Set Theory, he proves a quite useful theorem (3.11) that I'm using as a vital part of a project I'm doing. However, there's a part of the proof I can't wrap my head around.
Let X be a completely metrizable space with a compatible metric d, and Y a $G_\delta$ subset of X. As such, Y can be written as $\cap_n U_n$, where the $U_n$ are open in X. Let $F_n = X\setminus U_n$, and define the metric d' on Y as such:
$d'(x,y) = d(x,y) + \sum_{n=0}^{\infty}min\left\{2^{-n-1}, \left|\frac{1}{d(x,F_n)} – \frac{1}{d(y,F_n)}\right|\right\}$.
Kechris then states that it is easy to show that d' generates the same topology as d. This is not obvious to me. The inclusion one direction is obvious, since $d(x,y)\leq d'(x,y)$ for all x,y, but I'm not seeing how to prove the other direction at all.
Any advice would be appreciated!
Best Answer
We have to show that if $y \in Y$ and $(x_m)$ is a sequence in $Y$ such that $d(x_m,y) \to 0$, then $d'(x_m,y) \to 0$. So let $\varepsilon > 0$. For $m \ge m_0$ we have $d(x_m,y) < \varepsilon/2$. Choose $N$ such that $\sum_{n=N+1}^\infty 2^{-n-1} < \varepsilon/4$. Hence it suffices to show that $\sum_{n=0}^N \left|\frac{1}{d(x_m,F_n)} - \frac{1}{d(y,F_n)}\right|< \varepsilon/4$ for $m \ge M$. To do so, it suffices to show that for each $n$ we have $\left|\frac{1}{d(x_m,F_n)} - \frac{1}{d(y,F_n)}\right| < \varepsilon/4(N+1)$ for $m \ge M_n$. Then take $M = \max(M_0,\dots,M_N)$. Now observe that the distance function $d(-,F_n)$ is continuous so that $\left|\frac{1}{d(x_m,F_n)} - \frac{1}{d(y,F_n)}\right| = \left|\frac{d(y,F_n) - d(x_m,F_n)}{d(x_m,F_n)d(y,F_n)}\right| \to \frac{0}{d(y,F_n)^2} = 0$.