Real analysis- open and closed set I just signed up on this website because I can't find the way to solve these questions anywhere.

Let $( ℝ , d)$ be a metric space , where $d(x,y)=|x-y|$ , $∀ x,y ∈ ℝ$.
Show that the singleton $\{x\}$ is closed set.
A metric space is often imagined as a box with a lot of points in it.



Where each pair of points has a distance assigned to it $d(x,y)$. The distance is assumed to be positive and only zero when $x=y$(aka if you have different points their distance is positive).


Along with a notion of distance comes a notion of neighboorhoods or open balls. An open ball $B(x,r)$ is simply the set of points $z$ in the space that have distance $d(x,z)$ less than $r$. 

An open set is a set $V$ with which for every point $x$ in $V$ you can find an open ball of $x$ contained inside of $V$.

Closed sets $C$ however are defined to be the sets that contain all of their limit points. In the picture the black area around C is also meant to be apart of it. Limit points of $C$ are points that always have an open ball "touching" $C$. So if a point isn't a limit point it has an open ball that doesn't touch $C$. 

Also note that since every limit point is in $C$, any point not in $\color{red}{C}$ () or is in $\color{blue}{R/C}$ has an $\color{purple}{\textrm{open ball}}$ not "touching" $\color{red}{C}$


With all that said. Assume we don't know that $\{x\}$ is closed but want to show that it is closed (I have drawn the set $\{x\}$ as a white circle containing $x$). We need to show that every point $y$ $\color{blue}{\textrm{outside}}$ of $\{x\}$ has an $\color{purple}{\textrm{open ball}}$ that doesn't "touch" $\{x\}$. Since for any singular point $y$ outside $\{x\}$ we know the distance between $x$ and $y$ we simple make an $\color{purple}{\textrm{open ball}}$ around $y$ with radius less than half the distance between $x$ and $y$.
Is it correct ? @Student If not what to change?