A basic question on limit point of an infinite subset Suppose in a metric space $X$ for a given $\delta >0$ we can find an infinite subset of $X$ in which the distance between any two elements is $\geq \delta$. Then can we say anything about whether the set has a limit point or not ?  
 A: Let $A$ be such a set. If $x\in A$, then $N(x,\delta)\cap A=\{x\}$, so $x$ is not a limit point of $A$. If $x\in X\setminus A$, there are two possibilities. If there is an $a\in A$ such that $d(x,a)<\delta$, let $\epsilon=d(x,a)$; then $N(x,\epsilon)\cap A=\varnothing$, so $x$ is not a limit point of $A$. And if there is no such $a\in A$, then $N(x,\delta)\cap A=\varnothing$, and again $x$ is not a limit point of $A$.
We can also conclude that $X$ is not compact: $A$ is closed, so $$\mathscr{U}=\{N(x,\delta):x\in A\}\cup\{X\setminus A\}$$ is an open cover of $X$ with no finite subcover, since if $x\in A$, $N(x,\delta)$ is the only member of $\mathscr{U}$ containing $x$.
A: Let $X$, $\delta$, and $S$ be as described. Suppose $x\in X$ is a limit point of $S$. Then there must be a point $a\in S\setminus\{x\}$ such that $d(x,a)<\frac\delta2$ and a point $b\in S\setminus\{x\}$ such that $d(x,b)<d(x,a)<\frac\delta2$. Then $b\ne a$ because $d(x,b)<d(x,a)$ and by the triangle inequality, $d(a,b)\le d(x,a)+d(x,b)<\delta$, a contradiction.
