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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,881
Hobart, Tasmania
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:



Willard - 1 - Theorem 3.7 ... PART 1 ... .png
Willard - 2 - Theorem 3.7 ... PART 2 ... .png



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
Feb 7, 2012
2,648
Leeds, UK
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. :)