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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
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:



Willard - Interior ... Defn 3.9, Lemma 3.10 and Theorem 3.11 .png


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:



Willard - Defn 3.5, Lemma 3..6 and Theorem 3.7 .png



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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Last edited:

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
338
Brazil
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
Jun 22, 2012
2,886
Hobart, Tasmania
Thanks for that Fantini ... I'm well ...

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

Peter
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
338
Brazil
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
Jun 22, 2012
2,886
Hobart, Tasmania
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
 
Last edited:

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
338
Brazil
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!

Regards,

Fantini.