# closure of sets (topology)

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• Feb 17th 2009, 06:21 AM
tomboi03
closure of sets (topology)
i don't know how to prove this..
if it's false then i'm suppose give an example where the equality fails...

So, the closure of the intersection of all Aalpha = intersection of all closure of Aalpha.

the closure of the union of all Aalpha = union of all closure of Aalpha.

I know for the last one it is false... but i can't find an example...

Please help me out!

Thank You!
• Feb 17th 2009, 06:38 AM
HallsofIvy
﻿
Quote:

Originally Posted by tomboi03
i don't know how to prove this..
if it's false then i'm suppose give an example where the equality fails...

So, the closure of the intersection of all Aalpha = intersection of all closure of
Aalpha.

the closure of the union of all Aalpha = union of all closure of Aalpha.

I know for the last one it is false... but i can't find an example...

Please help me out!

Thank You!

The first statement is true but how you prove it depends entirely on how you DEFINE "closed
set" and there are several equivalent ways. For example, it is possible to define "closed set" as
being a member of a collection of all subset of the universal set U such that
1) U is in the collection
2) The empty set is in the collection
3) The intersection of any family of sets in the collection is in the collection
4) The union of any finite family of sets in the collection is in the collection.
In this case, you theorem is just (3) in the definition of "closed"! (This particular definition is not
often used.)

What definition of "closed set" are you using?

As for the second, yes it is true and you only need to give a single counter-example. Consider
$\displaystyle A_n= \left[\frac{1}{n}, 1- \frac{1}{n}\right]$ for all integers n> 1. That is
$\displaystyle A_2= \left[\frac{1}{2},1-\frac{1}{2}\right]= \{\frac{1}{2}\}$,
$\displaystyle A_3=\left[\frac{1}{3},1-\frac{1}{3}\right]= \left[\frac{1}{3}, \frac{2}{3}\right]$,
$\displaystyle A_4=\left[\frac{1}{4},1-\frac{1}{4}\right]= \left[\frac{1}{4}, \frac{3}{4}\right]$
You will need to show that each of those sets is closed, that the union of all such sets is (0, 1),
and that (0, 1) is not closed.
(Note that proving it is open is NOT proving it is closed.)

Here [a,b] is the closed interval and (a, b) is the open interval in the real numbers.