0
$\begingroup$

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 $A\subseteq \overline {A}$.

$\endgroup$
  • $\begingroup$ What is your definition of $\overline{A}$? $\endgroup$ – Jack Jan 14 '17 at 21:42
  • $\begingroup$ @Jack Edited... $\endgroup$ – PozcuKushimotoStreet Jan 15 '17 at 0:54
  • 1
    $\begingroup$ 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. $\endgroup$ – user 170039 Jan 15 '17 at 4:05
  • $\begingroup$ Wait so are you using the proposition? $\endgroup$ – Jacob Wakem Jan 15 '17 at 4:24
  • 1
    $\begingroup$ By the definition you gave, you don't need the proposition at all. $\endgroup$ – Jack Jan 15 '17 at 16:46
3
$\begingroup$

This property holds trivially by definition. Indeed, $\overline{A}$ is the smallest closed set containing $A$.

$\endgroup$
  • 5
    $\begingroup$ it somewhat depends on your definition, it is unlikely that op is using this definition $\endgroup$ – qbert Jan 14 '17 at 21:34
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

If $x\in A$ then $x\in B$ for every closed subset $B$ containing $A$. Therefore $x\in\overline A$ by (your) definition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.