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.
To start with, let us reinforce the definition of adherent point.
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 ($\Rightarrow$).
At this point, I got stuck. Could someone please help me how to solve it?
I also would like to know if such results keep valid in arbitrary metric spaces.
Any comments on the wording of my solution is welcome.
 A: 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 $X$.
