Open subset of a polish space

Question

HERE is the theorem about being polish of open subsets of polish spaces. Let $X$ be a Polish space and the complete metric on $X$ is $d<1$ and the new one defined on any open set $U\subset X$ is as $d^{*}(x,y)=d(x,y)+ |\frac{1}{d(x,X\setminus U)}-\frac{1}{d(y,X\setminus U)}|$ clearly it is a metric on $U$ so it says that for each $i$, $\frac{1}{d(x_{i},X\setminus U)}$ is greater than $1$, so bounded away from zero, how is that $d(x, X\setminus U)>0$ ($x$ is the limit of $x_{n}$ in $X$.

Answer 1

For $k\in\Bbb N$ let

$$r_k=\frac1{d(x_k,X\setminus U)}\,;$$

You know that $\langle r_k:k\in\Bbb N\rangle$ is a Cauchy sequence in $\Bbb R$ (with the usual metric), and $\Bbb R$ is complete, so this sequence converges to some $r\in\Bbb R$. And $d(x_k,X\setminus U)<1$ for each $k\in\Bbb N$, so $r_k=\frac1{d(x_k,X\setminus U)}>1$ for each $k\in\Bbb N$. This immediately implies that $r\ge 1$: if $r$ were less than $1$, $(r-1,1)$ would be an open nbhd of $r$ not containing any $r_k$, and the sequence $\langle r_k:k\in\Bbb N\rangle$ would not converge to $r$.

Now $\langle x_k:k\in\Bbb N\rangle$ converges to $x$, and $d$ is continuous, so $\langle d(x_k,X\setminus U):k\in\Bbb N\rangle$ converges to $d(x,X\setminus U)$, and therefore $\langle r_k:k\in\Bbb N\rangle$ converges to $\frac1{d(x,X\setminus U)}$. Thus,

$$\frac1{d(x,X\setminus U)}=r\ge 1\,,$$

and

$$d(x,X\setminus U)=\frac1r>0\,.$$