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Boundary of Subset in a Topological Space ... Singh Theorem 1.3.10 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
I am reading Tej Bahadur Singh: Elements of Topology, CRC Press, 2013 ... ... and am currently focused on Chapter 1, Section 1.2: Topological Spaces ...

I need help in order to fully understand Singh's proof of Theorem 1.3.10 ...


Theorem 1.3.10 (plus the definition of boundary) reads as follows:



Singh - Defn 1.3.9 and Theorem 1.3.10 ... .png



In the above proof by Singh we read the following:

" ... ... Conversely, if \(\displaystyle x \in \overline{A} - A\), then \(\displaystyle x \in \overline{A} \cap (X - A) \subseteq \partial A\) and the reverse inclusion follows. ... ... "


My questions are as follows:


Question 1

Why is it true that \(\displaystyle \overline{A} \cap (X - A) \subseteq \partial A\) ... ?

I suspect that this is because \(\displaystyle (X - A) \subseteq \overline{ (X - A) }\) ... ... is that correct ... ?



Question 2

How does \(\displaystyle x \in \overline{A} \cap (X - A) \subseteq \partial A\) lead to the reverse inclusion being true ... ... ?

(I am assuming that the reverse inclusion is \(\displaystyle \overline{A} \subseteq A \cup \partial A\) ... )


Help will be much appreciated ... ...

Peter
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,485
Question 1

Why is it true that \(\displaystyle \overline{A} \cap (X - A) \subseteq \partial A\) ... ?

I suspect that this is because \(\displaystyle (X - A) \subseteq \overline{ (X - A) }\) ... ... is that correct ... ?
Yes, your suggestion is correct.

Question 2

How does \(\displaystyle x \in \overline{A} \cap (X - A) \subseteq \partial A\) lead to the reverse inclusion being true ... ... ?

(I am assuming that the reverse inclusion is \(\displaystyle \overline{A} \subseteq A \cup \partial A\) ... )
We need to show that $x\in \bar{A}=A\cup(\bar{A}-A)$ implies $x\in A\cup\partial A$. This requires considering two cases: $x\in A$ and $x\in \bar{A}-A$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
Yes, your suggestion is correct.

We need to show that $x\in \bar{A}=A\cup(\bar{A}-A)$ implies $x\in A\cup\partial A$. This requires considering two cases: $x\in A$ and $x\in \bar{A}-A$.




Thanks Evgeny ...

Your post turned out to be really helpful ...

Appreciate your help ...

Peter