# Show that every adherent point of $X$ is either a limit point or an isolated point of $X$, but cannot be both.

Let $$X\subseteq\Bbb R$$ and $$x\in\Bbb R$$. Show that every adherent point of $$X$$ is either a limit point or an isolated point of $$X$$, but cannot be both. Conversely, show that every limit point and every isolated point of $$X$$ is an adherent point of $$X$$.

MY ATTEMPT

Let us try to prove $$(\Leftarrow)$$ first.

Given $$X\subseteq\Bbb R$$, we say that $$x\in\Bbb R$$ is an adherent point of $$X$$ if $$\forall\varepsilon > 0$$ there corresponds a $$y\in X$$ s. t. $$|x-y|\leqslant \varepsilon$$.

If $$x$$ is a limit point of $$X$$, then it is an adherent point of $$X\setminus\{x\}$$. Thus there exists a sequence $$x_n\in X\setminus\{x\}$$ s. t. $$x_n\to x$$.

In other words, $$(\forall\varepsilon > 0)(\exists N_{\varepsilon}\in\Bbb N_0)$$ s. t. $$n\geqslant N_{\varepsilon}\implies 0 < |x_n - x| \leqslant\varepsilon$$.

In particular, $$\forall\varepsilon > 0$$ there is a term $$x_{N_{\varepsilon}}\in X$$ s. t. the definition of adherent point is satisfied.

Similarly, we say that $$x\in X$$ is an isolated point if there is an $$\varepsilon > 0$$ s. t. $$|x - y| > \varepsilon\ \forall y\in X\setminus\{x\}$$.

Nonetheless, $$x$$ is still an adherent point of $$X$$ in this case.

This is because no matter which $$\varepsilon > 0$$ one chooses, it suffices to pick $$y = x$$ and the relation $$|x - y| = 0 < \varepsilon$$ always holds.

Let us try to prove ($$\Rightarrow$$).

In the first part there is no need to introduce sequences. If $$x$$ is a limit point of $$X$$, then $$x$$ is an adherent point of $$X\setminus\{x\}$$, so for each $$\epsilon>0$$ there is a $$y\in X\setminus\{x\}$$ such that $$|x-y|\le\epsilon$$; certainly $$y\in X$$, so $$x$$ is an adherent point of $$X$$. And if $$x$$ is an isolated point of $$X$$, then $$x\in X$$, and of course $$|x-x|<\epsilon$$ for every $$\epsilon>0$$, so $$x$$ is an adherent point of $$X$$.
For the other direction I would show that if $$x$$ is an adherent point of $$X$$ that is not a limit point of $$X$$, then $$x$$ is an isolated point of $$X$$. If $$x$$ is not a limit point of $$X$$, there is an $$\epsilon>0$$ such that $$|x-y|>\epsilon$$ for each $$y\in X\setminus\{x\}$$. And if $$x$$ is an adherent point of $$X$$, then ...
Yes, this is true for all metric spaces. In a slightly more general form it is true for all topological spaces: it says that a point $$x$$ is in the closure of a set $$X$$ if and only if it is in the closure of $$X\setminus\{x\}$$ or is an isolated point of $$X$$.