# Show that $A\subseteq \overline {A}$

Definition. A subset of $\mathbb{R}$ is closed if it's complement is open.

Proposition. For any subset $A$ of $\mathbb{R}$ there is a unique subset $\overline {A}$ containing $A$ with the property that if $B$ is closed set containing $A$ then $\overline {A}\subseteq B$.

$\overline {A}=\bigcap${ $B\subseteq \mathbb{R}$ $B$ is closed and contains $A$}

Question. How can I show $A\subseteq \overline {A}$ to use the proposition? Can you give a hint?

My proof trying is: Let $A\subseteq \mathbb{R}$. Let $\overline {A}$ be closure of $A$. We will show that $A\subseteq \overline {A}$.

• What is your definition of $\overline{A}$? – Jack Jan 14 '17 at 21:42
• @Jack Edited... – PozcuKushimotoStreet Jan 15 '17 at 0:54
• If $A\not\subseteq \overline{A}$ then there exist $x\in A$ such that $x\not\in\overline{A}$. But this would mean that (by the definition of $\overline{A}$) for all closed set $B$ with $A\subseteq B$ we have $x\not\in B$. Since $A\subseteq B$ and $x\not\in B$ we must have $x\not\in A$ which contradicts our hypothesis. – user 170039 Jan 15 '17 at 4:05
• Wait so are you using the proposition? – Jacob Wakem Jan 15 '17 at 4:24
• By the definition you gave, you don't need the proposition at all. – Jack Jan 15 '17 at 16:46

This property holds trivially by definition. Indeed, $\overline{A}$ is the smallest closed set containing $A$.
I claim that $$\overline{A}= \bigcap_{i \in I} B_i,$$ where $I$ is just an index set and $B_i$ ranges over all closed sets that contain $A$. Here, $\overline{A}$ is clearly closed (by DeMorgan's laws, or definition) and if $B$ is closed and contains $A$, then it is among the $B_i$ , so $\overline{A} \subseteq B_i$.
To show uniqueness, just note that if another set $B$ has this property, $\overline{A} \subseteq B$ by construction and $B \subseteq \overline{A}$ by hypothesis.
If $x\in A$ then $x\in B$ for every closed subset $B$ containing $A$. Therefore $x\in\overline A$ by (your) definition.