# Let $X$ be a subset of $\mathbb R$. Show that $\bar X$ is closed

Let $X$ be a subset of $\mathbb R$. Show that $\bar X$ is closed (i.e, $\bar {\bar X} = \bar X$). Futhermore, show that if $Y$ is any enclosed set that contains $X$, then $Y$ also contains $\bar{X}$. Thus, the closure of $\bar X$ of $X$ is the smallest closed set which contains $X$.

## My Attempt

Proof

$\bar E =\bar {\bar E}$; one contains the other

If $x$ is an adherent point of $\bar E$, then $x$ is an adherent point of $E$. If $x$ is adherent point of $\bar E$.

=>For all $\epsilon>0$ there exists an $N$ for all $n \geq N |a_n-x| \leq \epsilon$

=>For all $\epsilon>0$ $|b_n-x|\leq |a_n - b_n| + |a_n - x| \leq \epsilon /2$

=> $b_n \rightarrow X, x \in E$

• What is your definition of the closure? What you're trying to prove is what I usually consider the definition. I've also never heard of an adherent point, but maybe that's just something I'm not familiar with. – Matt Samuel May 4 '16 at 3:22
• To elaborate on what Matt Samuel has said: usually, for $X \subseteq\mathbb R$, we define $\mathcal C_X = \{C \subseteq \mathbb R : X \subseteq C \text{ and } C \text{ is closed } \}$ and then define $$\overline X := \bigcap_{C \in \mathcal C_X} C.$$ That is, we define the closure of $X$ to be the intersection of all closed sets containing $X$. From this definition, it is immediately clear that $\overline X = \overline{\overline X}$ and that if $C$ is closed an contains $X$ then $C$ contains $\overline X$. – User8128 May 4 '16 at 4:02
• Proof critique: where does the sequence suddenly come from? What is your definition of adherent point? – Henno Brandsma May 4 '16 at 4:05

From your proof attempt, I see that you probably have the following definition of the closure of $E$ (which is a standard one):

$$\overline{E} = \{x \in \mathbb{R}: x \text{ adherent to } E\}$$

where $x$ is adherent to $E$ iff for all $\varepsilon>0$, there exists some $a \in E$ such that $|x - a| < \varepsilon$. (The last definition can be generalised to any metric space, or even any topological space as well, using open balls resp. open sets.)

It is clear that if $A \subseteq B$, then $\overline{A} \subseteq \overline{B}$, because the same point that would be used (for some $\varepsilon$) to show $x$ is adherent to $A$ also shows it's adherent to $B$. As clearly $E \subseteq \overline{E}$, this implies $\overline{E} \subseteq \overline{\overline{E}}$

So suppose $x$ is adherent to $\overline{E}$. Let $\varepsilon > 0$. For the number $\frac{\varepsilon}{2} > 0$ we can apply that we know that $x$ is adherent to $\overline{E}$, so there is some $b \in \overline{E}$ such that $|x- b| < {\varepsilon \over 2}$.

Now as $b$ is adherent to $E$, in turn we find a point $a \in E$ such that $|b -a | < {\varepsilon \over 2}$. So in all $|x - a| \le |x - b| + |b -a | < {\varepsilon \over 2} + {\varepsilon \over 2} = \varepsilon$, which shows that (as $\varepsilon$ was arbitrary) that $x \in \overline{E}$.

So we also have $\overline{\overline{E}} \subseteq \overline{E}$.

The ${\varepsilon \over 2}$-idea was sound, but the sequence is unnecessary.

The last statement is easy: if $X \subseteq Y$ and $Y$ is closed, then by the first remarks above we get that $\overline{X} \subseteq \overline{Y} = Y$, the last equality because $Y$ is closed. So any closed set that contains $X$ also contains its closure.