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.
Best Answer
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 $X$.