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

#### Peter

##### Well-known member
MHB Site Helper
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: 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
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 $$\displaystyle \overline{A} \subseteq A \cup \partial A$$ ... )
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

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 ...