# Closure and Interior as Dual Notions ... Proving Willard Theorem 3.11 using Duality ... ...

#### Peter

##### Well-known member
MHB Site Helper
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:

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

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#### Fantini

MHB Math Helper
Hello, Peter. It has been a while. How are you? I believe this is a solution:

For every $$x \in A^{\circ}$$ we have that there exists $$G \subset X$$ with $$x \in G$$. Remember $$G$$ is open and $$G \subset A$$, by the definition of $$A^{\circ}$$. Since $$G \subset A$$ it follows $$x \in A$$. Since we proved that $$x \in A^{\circ} \implies x \in A$$ for every $$x \in A^{\circ}$$ it implies $$A^{\circ} \subset A$$.

Regards,

Fantini.

#### Peter

##### Well-known member
MHB Site Helper
Thanks for that Fantini ... I'm well ...

BUT ... the problem was to prove it using duality with closure ...

Peter

#### Fantini

MHB Math Helper
I'm sorry. I wasn't clear on what you asked.

We have \begin{aligned} X - A^{\circ} & = X - \left( \bigcup \left\{ G \subset X \; | \; G \text{ open}, G \subset E \right\} \right) \\ & = \left( \bigcup \left\{ G \subset X \; | \; G \text{ open}, G \subset E \right\} \right)^{c} \\ & = \bigcap \left( \left\{ G \subset X \; | \; G \text{ open}, G \subset E \right\} \right)^{c} \\ & = \bigcap \left\{ G^{c} \subset X \; | \; G^{c} \text{ open}, E^c \subset G^c \right\} \\ & = \bigcap \left\{ K \subset X \; | \; K \text{ closed}, X-E \subset K \right\} \\ & = \overline{X-E} \end{aligned} I've adopted the following notation: $$M^c = X-M$$ to lessen the burden. It is worthy to note that:
• from line 1 to 2 I've merely changed the notation;
• from line 2 to 3 I've used DeMorgan's Laws;
• from line 3 to 4 I've used DeMorgan's Laws as well;
• from line 4 to 5 I reversed the notation back.
I believe the proof of $$X - \overline{A} = \left( X-A \right)^{\circ}$$ would go the same way.

Regards,

Fantini.

#### Peter

##### Well-known member
MHB Site Helper
Thanks Fantini ...

I really appreciate your help ... and I apologize for not making my question clear ...

The problem was ... given Theorem 3.7 K-a ... namely $$\displaystyle E \subset \overline{E}$$ for any subset $$\displaystyle E$$ of a topological space $$\displaystyle X$$ ... prove Theorem 3.11 1-a ... namely $$\displaystyle A^{ \circ } \subset A$$ for any subset $$\displaystyle A$$ of a topological space $$\displaystyle X$$ ... using only the definitions of closure and interior and the duality relations between closure and interior, namely $X - A^{ \circ } = \overline{ X - A }$ and $X - \overline{ A} = ( X - A)^{ \circ }$ ...

After some refection (and a hint given to me on the PF) I formulated the following proof:

To show that given $$\displaystyle E \subset \overline{E}$$ for any subset $$\displaystyle E$$ of a topological space $$\displaystyle X$$ we can prove $$\displaystyle A^{ \circ } \subset A$$ for any subset $$\displaystyle A$$ of a topological space $$\displaystyle X$$ ...

Proof:

Since $$\displaystyle E \subset \overline{E}$$ for any subset $$\displaystyle E$$ of a topological space $$\displaystyle X$$ we have $$\displaystyle X - A \subset \overline{ X - A }$$

But ... $$\displaystyle \overline{ X - A } = X - A^{ \circ }$$

So $$\displaystyle X - A \subset X - A^{ \circ }$$ ...

... which implies ...

$$\displaystyle A^{ \circ } \subset A$$

Is that correct ... ?

Peter

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#### Fantini

Yes, if you intend to prove $$E \subset \overline{E} \implies A^{\circ} \subset A$$. I'm sorry I have twice missed the point. Good job!