Proving Willard theorem 3.11 part 1-a using closure-interior duality

[Math Amateur]

TL;DR

I am reading Stephen Willard: General Topology ... ... and am studying Chapter 2: Topological Spaces and am currently focused on Section 3: Fundamental Concepts ... ...

I need help in order to prove Theorem 3.11 Part 1-a using the duality relations between closure and interior ... ..

I am reading Stephen Willard: General Topology ... ... and am studying Chapter 2: Topological Spaces and am currently focused on Section 3: Fundamental Concepts ... ...

I need help in order to prove Theorem 3.11 Part 1-a using the duality relations between closure and interior ... ..The definition of interior and Theorem 3.11 read as follows:

Readers of this post necessarily need access to the "dual" theorem ... namely Theorem 3.7 ...

Theorem 3.7 (together with Willard's definition of closure and a relevant lemma) reads as follows:

So ... I need help in order to prove Theorem 3.11 1-a assuming the dual result in Theorem 3.7 ( that is K-a or $A \subset \overline{A}$ ) using only the definitions of closure and interior and the dual relations: $X - A^{ \circ } = \overline{ X - A }$ and $X - \overline{ A} = ( X - A)^{ \circ }$ ...

My attempt so far is as follows:

To show $A^{ \circ } \subset A$ ...

Proof:

Assume $A \subset \overline{ A}$ ..

Now we have that ...

$A \subset \overline{ A}$

$\Longrightarrow X - \overline{ A} \subset X - A$

$\Longrightarrow (X - A)^{ \circ } \subset X - A$ ...But how do I proceed from here ... ?Help will be much appreciated ... ...

Peter

 

[Infrared]

Try your argument again, but starting with $X-A\subset \overline{X-A}$ instead of with $A\subset\overline{A}$.

But there's no need to use any previous theorems here: by your definition, $A^{\circ}$ is a union of sets all of which are subsets of $A$, so $A^{\circ}$ is a subset of $A$.

 

[Math Amateur]

Thanks ...

I understand there is no need to use previous theorems ... just wanted to understand how duality between closure and interior worked ...

Will try your suggestion ...

Thanks again ...

Peter