How does closure equality imply subset in Willard theorem 3.7?

[Math Amateur]

TL;DR

I need help in order to understand Willard Theorem 3.7 concerning topological closure ...

I am reading Stephen Willard: General Topology ... ... and am currently reading 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 $A \subset B$, then by K-c, $\overline{B} = \overline{A} \cup \overline{ (B -A) }$ so that $\overline{A} \subset \overline{B}$ ... ... "Can someone please demonstrate, formally and rigorously, how $\overline{B} = \overline{A} \cup \overline{ (B -A) }$ implies that $\overline{A} \subset \overline{B}$ ...
Help will be much appreciated ...

Peter

 

[Math Amateur]

Wow ... that was quick ...

Yes that's right ... hmm is it as simple as that ...

Thanks ...

Peter

 

[member 587159]

Hi. Something went wrong with wrong with the formatting. I think it follows from $X\subseteq X\cup Y$ for sets $X,Y$.

(I deleted previous post, but it was correct).

 

[Math Amateur]

Thanks ...

I think I have corrected the formatting ...

I think you're correct ...

Thanks again ...

Peter

 

[Math Amateur]

Hi. Something went wrong with wrong with the formatting. I think it follows from $X\subseteq X\cup Y$ for sets $X,Y$.

(I deleted previous post, but it was correct).

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

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

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

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

Peter

 

[member 587159]

I would just write it like this:

$\overline{A}\subseteq \overline{A} \cup \overline{B - A} = \overline{B}$