Show that for each $\delta_1>0$ there is $\delta_2>0$ such that for $x$ in $X$ with $d(x,X’)\ge\delta_1$ we have $d(x,X-\{x\})>\delta_2.$

Question

Question: Let $(X,d)$ be a metric space such that $X’$, the set of all accumulation points is compact and for each $\epsilon>0$ the set $X-B(X’,\epsilon)$ is uniformly discrete. Show that for each $\delta_1>0$ there is $\delta_2>0$ such that for $x$ in $X$ with $d(x,X’)\ge\delta_1$ we have $d(x,X-\{x\})>\delta_2.$

For the last few days I was trying to solve it. Mainly I used the method of contradiction but could not reach at a solution.

Please help me.

Answer 1

Suppose not. Then there were a $\delta_1 \ge 0$ such that for all $n \in \mathbb N$, there are $x_n \ne y_n \in X$ with $d(x_n,X') \ge \delta_1$ and $d(x_n,y_n) \le \frac 1n$. By passing to a subsequence, we may assume that $d(y_n, X') \ge \frac{\delta_1}2$ for all $n \in \mathbb N$. But $Y := X - B(X',\frac{\delta_1}2)$ is uniformly discrete, contradicting $d(x_n, y_n) \le \frac 1n \to 0$.