How to show that a simple set is closed? I am comfortable with the intuition of closed sets, however, I am struggling to formalise these ideas.
Given the set $X =$ { $(x,0)$ | $x \le 0$ and $x \in \mathbb{R}$}, I want to know how I can show that this set is closed.
Formally, to do this, we must show that if we take some arbitrary convergent sequence of points in $X$, then the limit must also be an element of $X$. This set seems like a fairly simplistic set to apply this to, but I am not entirely comfortable with constructing these types of arguments, and I'm unclear how to show this generally for some arbitrary sequence.
I would be grateful for any guidance or references.
 A: Suppose $\{x_n\}$ is a convergent sequence of points in $X$ which is to say they are real non-positive points.  As the converge let them converge to $x\in \mathbb R$.  All we have to do is to show that $x \in X$ or in other words that $x \le 0$.
So proving $X$ is closed is entirely equivalent to proving: every convergent sequence of non-positive number converge to a non-positive limit.
And that should be a familiar exercise.
Proof by contradiction:  Let $x_n\to x$ and $x_n \le 0$.  Now suppose $x > 0$.  Then if we let $\epsilon$ be so that $0 < \epsilon < x$ then for every $x_n$ we have $|x_n -x| \ge x > \epsilon$ and thus the definition of $x_n \to x$ fails.
Thus ..... every convergent sequence of reals less than or equal to zero converge to a real that is less than or equal to zero.  Or in other words, every convergent sequence of points in $X$ converge to a point in $X$ which is your definition of closed.
....
FWIW I do prefer the limit point definition.  It's equivalent, but somehow it is more intuitive to me.
That proof would go like this.
Let $s$ be a limit point of $s$.  Thus for every $\epsilon > 0$ we will have $N_(s,\epsilon)$ will have a point of $X$. In other words, for every $\epsilon > 0$ there will be a point $x$ so that $|x-s| < \epsilon$.
That means $x-\epsilon < s < x + \epsilon$.  But $x \le 0$ so $s < x+ \epsilon \le \epsilon$ for all $\epsilon$.  That meanx $s \le \epsilon$ for all positive $\epsilon$ and that is only possible if $s\le 0$ so $s\in X$.
.....
If that's too abstract:
Let $s \not \in X$.  Then $s > 0$.  Let $0 < \epsilon < s$ and let $x$ be any element of $X$.  Then $|x-s| > \epsilon$.  So $N(s,\epsilon)$ does not contain any points of $X$.
So $s$ is not a limit point of $X$.  So all the limit points of $X$ must be less than or equal to $0$ and thus by in $X$.
A: The space $\mathbf{R}^{2}$ is a metric space. Remark that a point $p$ is a limit point of set $X\subseteq \mathbf{R}^{2}$ if every neighborhood of $p$ contains a point $q\not=p$ such that $q\in X$ and $X$ is closed if every limit point of $X$ is a point of $X$.
Now see the following illustration.

