# Prove cl(int(cl(int(A))))=cl(int(A))

## Homework Statement

I am working on the proof that taking closure and interior of a set in a metric space can produce at most 7 sets. The piece I need is that $\bar{\mathring{A}} = \bar{\mathring{\bar{\mathring{A}}}}$.

## Homework Equations

Interior of A is the union of all open sets contained in A, aka the largest open set contained in A.
Closure of A is the intersection of all closed sets containing A, aka the smallest closed set containing A.

## The Attempt at a Solution

$\bar{\mathring{A}}$ is a closed set. $\mathring{\bar{\mathring{A}}}\subseteq \bar{\mathring{A}}$. Since $\bar{\mathring{\bar{\mathring{A}}}}$ is the smallest closed set containing $\mathring{\bar{\mathring{A}}}$ we have that $\bar{\mathring{\bar{\mathring{A}}}}\subseteq \bar{\mathring{A}}$.

I'm not sure how to get the inclusion $\bar{\mathring{A}}\subseteq \bar{\mathring{\bar{\mathring{A}}}}$

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Dick
Homework Helper
int(A) is an open set contained in cl(int(A)). What does this tell you about its relation to int(cl(int(A)))?

Last edited:
int(A) is an open set contained in cl(int(A)). What does this tell you about it's relation to int(cl(int(A)))?

$\mathring{A}\subseteq \mathring{\bar{\mathring{A}}}$

Dick
$\mathring{A}\subseteq \mathring{\bar{\mathring{A}}}$