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 $X$.


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?


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.

  • $\begingroup$ When you are proving your first part, you write "Since $\overline{X}$ is closed". Isn't that precisely what you wish to show? You cannot use that to argue. Also, it could be useful if you mention what are the definitions that you are using. (For example, some define $\bar{X}$ to be the smallest closed set containing $X$.) $\endgroup$ – Aryaman Maithani May 22 '20 at 21:20
  • $\begingroup$ Thanks for the comments. I have just fixed it. Hopefully it is correct now. $\endgroup$ – BrickByBrick May 22 '20 at 21:44
  • $\begingroup$ The solution now looks OK to me. $\endgroup$ – Aryaman Maithani May 22 '20 at 21:50

Yes, using the adherent point definition, this solution is correct, IMO. The second part is also fine.

Minor grammatical/logical point: start with an arbitrary $\varepsilon>0$. The split in half (so the fact that you actually apply the definition of adherent point for $\frac{\varepsilon}{2}$ (twice)) is your own addition to use the triangle inequality. The end result is that $B(x, \varepsilon)$ intersects $X$ and then you use that $\varepsilon>0$ was arbitrary to conclude $x \in \overline{X}$ etc.


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