Let $X$ be a subset of $\textbf{R}$. Show that $\overline{X}$ is closed. Furthermore, show that if $Y$ is any closed set that contains $X$, then $Y$ also contains $\overline{X}$. Thus the closure $\overline{X}$ of $X$ is the smallest closed set which contains $X$.
MY ATTEMPT
Let $x\in\textbf{R}$ be an adherent point of $\overline{X}$. Then for every $\varepsilon/2 > 0$, there is an $y\in\overline{X}$ such that $|x-y|\leq\varepsilon/2$. But $y\in\overline{X}$ is an adherent point of $X$. Consequently, for the same $\varepsilon/2 > 0$ there corresponds an $z\in X$ such that we have that $|y - z| \leq \varepsilon/2$.
Hence we conclude that $x\in\overline{X}$. Indeed, one has that \begin{align*} |x - z| \leq |x - y| + |y - z| < \varepsilon \end{align*}
We have just proven that $\overline{X}\supseteq\overline{\overline{X}}$. Since the converse inclusion $\overline{\overline{X}}\supseteq\overline{X}$ does always hold, the result follows.
If $Y$ is closed, then $Y = \overline{Y}$. Making use of the properties of closure, we have that \begin{align*} X\subseteq Y \Rightarrow \overline{X}\subseteq\overline{Y} = Y \end{align*} and the result holds.
Is my wording of the proof formal enough?
Comments
The definition of closure which I was given is the following:
The closure of a set $X\subseteq\textbf{R}$ is the set of all its adherent points.
