Equivalence of isolated point in metric space and singleton open set in topological space

Question

In mathematics, a point $x$ is called an isolated point of a subset $S$ (in a topological space $X$) if $x$ is an element of $S$ but there exists a neighborhood of $x$ which does not contain any other points of $S$. This is equivalent to saying that the singleton $\{x\}$ is an open set in the topological space $S$ (considered as a subspace of $X$).

I found this here, but i really could not understand the equivalence of the two concepts. Some good explanation would be a great help as I am new to topological spaces.

Answer 1

Suppose that $\{x\}$ is a neighborhood of $x$. Then there is an open set $O$ such that $x\in O$ and that $O\subset\{x\}$. But then $O=\{x\}$ and therefore $\{x\}$ is open.

On the other hand, if $\{x\}$ is open, then $\{x\}$ is a neighborood of $x$ which has $x$ as its only element.

Answer 2

We have a subset $S$ of a topological space $X$. We want to prove that $x\in S$ is isolated in $S$ if and only if $\{x\}$ is open in $S$ with the subspace topology.

Recall that a set $A\subset S$ is open in the subspace topology means that $A=A'\cap S$ where $A'\subset X$ is open.

For the forward direction, if $x$ is isolated, then there is a neighborhood $A'\subset X$ of $x$ which is open such that $x$ is the only point in $S$ also contained in $X$. This means that $A=A'\cap S$ is open in the subspace topology on $S$, but $A'\cap S=\{x\}$, so $\{x\}$ is open.

For the backward direction, if $\{x\}$ is open in the subspace topology on $S$, that means that $\{x\}=A'\cap S$ for some open subset $A'$ of $X$. This means that $A'$ is a neighborhood of $x$ which does not contain any other points of $S$, so $x$ is isolated.