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.

Please see https://math.meta.stackexchange.com/questions/5020/ — Angina Seng, Nov 01 '18 at 06:20
There is a positive $\epsilon$ such that $d(a,b)>\epsilon$ for all $a, b (a\ne b)$ in that set — Jave, Nov 01 '18 at 06:39
Are you satisfied with the link given above? What issue do you have with it, for instance? — Sarvesh Ravichandran Iyer, Nov 01 '18 at 06:46

score 0 · Accepted Answer · answered Nov 01 '18 at 07:47

0

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 $d(x_n, y_n) \le \frac 1n \to 0$.

answered Nov 01 '18 at 07:47

martini

84,101

Thank you so much – Jave Nov 01 '18 at 16:21

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 $d(x,X-\{x\})>\delta_2.$

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