A few days ago, I asked this question (Prove that every infinite metric space $(X, d)$ contains an infinite subset $A$ such that $(A, d)$ is discrete.).

Thanks for egreg's answer: Suppose $X$ is not discrete (otherwise you're already done).

Then there is $a\in X$ which is not an isolated point; so, for every $n>0$, there is a point $x_n\in X$, $x_n\ne a$, such that $d(x_n,a)<1/n$.

(It is not difficult to build the sequence so that, for every $m$, $x_1,x_2,\dots,x_m$ are pairwise distinct, but it's not really required.)

Consider the set $A=\{x_n:n>0\}$. Then $A$ is infinite and has no limit point, because…

I have tried very hard to understand the answer. But it seems that there is just no explicit construction there, without which I cannot prove each point $x_n$ is isolated. May I please ask for an explicit way of choosing $x_n$ so that an appropriate radius can be chosen for each $x_n$ such that the open ball centred $x_n$ with that radius contains nothing from $A$ but $x_n$?


Suppose otherwise that $x_n$ is not isolated. Then there is a sequence of points from $A$ that converges to $x_n$; in particular, infinitely many points $x \in A$ are within distance $d(x_n,a)/2$ of $x_n$ (i.e. $d(x,x_n) \le d(a,x_n)/2$). But then, these infinitely many points $x \in A$ satisfy $$d(x,a) \ge d(x_n,a) - d(x,x_n)\ge d(x_n,a)/2,$$ which contradicts the construction of the original sequence $A$ converging to $a$.

  • $\begingroup$ May I please ask: Why is there a contradiction? A point is a limit point of $A$ iff any open ball centred at that point contains at least a point from $A$ which is different from that centre point. There is no sequence convergence involved. $\endgroup$ – PropositionX Mar 24 '17 at 4:07
  • $\begingroup$ @Y.X. The original sequence $x_1,x_2,\ldots$ constructed in egreg's answer converges to $a$, by the $d(x_n,a)<1/n$ condition. The contradiction is that you cannot have infinitely many of these $x_m$ satisfying $d(x_m,a) \ge r$ for some $r>0$. $\endgroup$ – angryavian Mar 24 '17 at 4:32

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