# Closure in a Topological Space ... Willard, Theorem 3.7 ... ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Stephen Willard: General Topology ... ... and am currently focused on Chapter 2: Topological Spaces and am currently focused on Section 1: Fundamental Concepts ... ...

I need help in order to fully understand an aspect of the proof of Theorem 3.7 ... ..

Theorem 3.7 and its proof read as follows:

In the above proof by Willard we read the following:

" ... ... First note that if $$\displaystyle A \subset B$$, then by K-c, $$\displaystyle \overline{B} = \overline{A} \cup \overline{ (B -A) }$$ so that $$\displaystyle \overline{A} \subset \overline{B}$$ ... ... "

Can someone please demonstrate, formally and rigorously, how $$\displaystyle \overline{B} = \overline{A} \cup \overline{ (B -A) }$$ implies that $$\displaystyle \overline{A} \subset \overline{B}$$ ......

Help will be much appreciated ...

Peter

EDIT: it might be as simple as ... for two sets $$\displaystyle X$$ and $$\displaystyle Y$$ we have $$\displaystyle X \subset X \cup Y$$ ... ... is that correct?

Peter

EDIT 2:
I am probably being pedantic ... but it might be more accurate to say that ...

$$\displaystyle \overline{B} = \overline{A} \cup \overline{ (B -A) }$$ implies that $$\displaystyle \overline{A} \subset \overline{B}$$ ... ... is true because ...

... for sets $$\displaystyle X, Y$$ and $$\displaystyle Z$$ we have that ...

$$\displaystyle X = Y \cup Z \Longrightarrow Y \subset X$$... ...

Peter

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
EDIT: it might be as simple as ... for two sets $$\displaystyle X$$ and $$\displaystyle Y$$ we have $$\displaystyle X \subset X \cup Y$$ ... ... is that correct?
Yes, that is correct.