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

1. Oct 19, 2013

### ArcanaNoir

1. The problem statement, all variables and given/known data

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 $\bar{\mathring{A}} = \bar{\mathring{\bar{\mathring{A}}}}$.

2. Relevant 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.

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

Last edited: Oct 19, 2013
2. Oct 19, 2013

### Dick

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: Oct 19, 2013
3. Oct 19, 2013

### ArcanaNoir

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

4. Oct 19, 2013

### Dick

Ok, so it's pretty easy to finish from there, right?

5. Oct 19, 2013

### ArcanaNoir

oh! Thank you :)