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: 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 $y$.

A: We want to prove that $\{x\}$ is a closed subset of the metric space $(\mathbb{R}, d)$. This is equivalent with proving that the complement (= $\mathbb{R} - \{x\}$) is an open set. 
In order to prove that $\mathbb{R} - \{x\}$ is open, consider an element $y \in \mathbb{R} - \{x\}$. In particular, we have that $x \neq y$. We want to find an open ball of some radius $r$ with center $y$, say $B(y, r)$ such that $B(y,r) \subseteq \mathbb{R} - \{x\}$. In particular, this means that $x \not\in B(y,r)$ which is equivalent with $d(x,y) \geq r$.
Since you have a metric, we can measure the distance between $y$ and $x$ and since $x\neq y$, this distance will be strictly larger than $0$, that is
$$d(x,y) > 0.$$ 
Now consider the open ball $B(y, \frac{d(x,y)}{2}) = \{z \in \mathbb{R} \vert d(y,z) < \frac{d(x,y)}{2}\}$. It is clear that $x \neq B(y,\frac{d(x,y)}{2})$ since $d(x,y) > \frac{d(x,y)}{2}$. Hence the set $\mathbb{R} - \{x\}$ is an open set, so $\{x\}$ is a closed set. 
$\textbf{REMARK}$: we could have considered $B(y, d(x,y))$ and we still would have that $x \not\in B(y, d(x,y))$, however, I really wanted to make it clear to you that you can find an open ball around $y$ not containing $x$.
A: In order to show the set $\{x\}$ is closed, one way to do is to prove the complement is open or to prove for each $y \notin \{x\}$, $\exists $ $\epsilon(y)$ such that $\{x\} \cap (y-\epsilon(y),y+\epsilon(y))= \phi$ .
So given any $y \in  \mathbb{R} \setminus \{x\} $, pick $\epsilon = d(x,y) > 0$, and so the open ball of radius $\epsilon$ centred at $y$ contained in $\mathbb{R} \setminus \{x\} $.
A: 
Is it correct ? @Student If not what to change? 
