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

[ArcanaNoir]

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}}}}

 

[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)))?

 

[ArcanaNoir]

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}}}

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

 

[ArcanaNoir]

oh! Thank you :)