Prove that $x \in \mathbb{R}$ is a limit point of a set $A \subset \mathbb{R}$ if and only if $d(x, A \setminus {x})=0$.

Question

Prove that $x \in \mathbb{R}$ is a limit point of a set $A \subset \mathbb{R}$ if and only if $d(x, A \setminus {x})=0$.

I think I have it right but I would like to have it checked.

We assume that $x$ is a limit point of the set $A$. Then we will have every neighbourhood $(x-r,x+r)$ where $r$ is the radius and is arbitrary intersects the set $A$, so we can phrase as $\forall r > 0 a \in A$ and $x \ne a$ such that $d(a,x)< r$. Now, $$\inf\{d(a,x) \text{ where } a \in A \text{ and } a \neq x\}=0. $$ Essentially letting $r$ become very small and close to $0$ to make $d(a,x)$ become $0$. Therefore, $d(x,A\setminus\{x\})=0$.

Now, we assume that $d(x,A\setminus\{x\})=0$. Because $d(x,A\setminus\{x\}) = \inf\{d(a,x) \text{ where } a \in A \text{ and } a \ne x\}$, we will have $$ \inf\{d(a,x) \text{ where } a \in A \text{ and } a \ne x\} =0. $$ Now, forall $r>0$ $a \in A$ and $x \ne a$ such that $d(a,x)< r$ which means every neighbourhood of $x$ contains a point of $A$ that is different from $x$. Therefore, $x$ is a limit point of the set $A$.

Answer 1

Improvement suggestion, staying close to the definition of infimum (no "letting $r$ get smaller and smaller" vagueness, but definitions):

If $x \in A'$ then let $s=d(x,A\setminus\{x\}) \ge 0$.

Suppose for a contradiction that $s >0$, and consider $B(x,s)$, which is an open ball around $x$, and as $x$ is a limit point of $A$, we can find $y \neq x$ such that $y \in B(x,s)$, or equivalently $d(x,y) < s$. But then $y \in A\setminus \{x\}$ and so $(s=)d(x, A\setminus\{x\}) \le d(x,y)(<s)$ (a lower bound of a set is $\le$ to each of its elements), but then $s < s$ which is a contradiction, so $s=0$ and indeed $d(x,A\setminus\{x\})=0$ as required.

Now suppose $d(x,A\setminus\{x\})=0$ and let $r>0$ be arbitrary. Then $r$ is not a lower bound of the set $D=\{d(x,y): y \in A\setminus\{x\}\}$, or it would be a strictly larger lower bound for $D$ which has largest lower bound $d(x,A\setminus\{x\})=0$. So some element of $D$ is smaller than $r$, or otherwise put, there is some $y \in A \setminus \{x\}$, so $y \in A, y \neq x$ such that $d(x,y) < r$. But this means this $y \in B(x,r)$ as required, and so $x$ is a limit point of $A$.

Answer 2

The metric $d$ is continuous. See here.

Now use the limit point definition of continuity, and a sequence $(x_n)\in A\setminus \{x\}$ with $x_n\to x$.

I think you have the right idea. But your wording is very awkward and not quite correct.