Let (X, O) be a topological space. Prove the following statements about the closed subsets of X:

(i) 0 and X are closed

(ii) If F1, F2 contained in X are closed, then the union of F1 and F2 is closed.

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- Jan 17th 2006, 07:40 AMTexasGirlProof in a topological space
Let (X, O) be a topological space. Prove the following statements about the closed subsets of X:

(i) 0 and X are closed

(ii) If F1, F2 contained in X are closed, then the union of F1 and F2 is closed. - Jan 17th 2006, 02:47 PMThePerfectHacker
Let $\displaystyle T$ be a topology on set $\displaystyle X$.

1) Since $\displaystyle \{\}\in T$ its complement is $\displaystyle X$ thus, $\displaystyle X$ is closed.

Now since $\displaystyle X\in T$ its complement is $\displaystyle \{\}$ thus, $\displaystyle \{\}$ is closed. - Jan 17th 2006, 02:54 PMThePerfectHacker
2)Since $\displaystyle F_1,F_2$ are closed there exists sets $\displaystyle S_1,S_2\in T$ respectively, such as their complenets give these sets. But, because $\displaystyle T$ is topological space then the union of any sets is an element of $\displaystyle T$ (by definition).But, the complement of $\displaystyle S_1\cup S_2$ is $\displaystyle F_1\cup F_2$. Thus, $\displaystyle F_1\cup F_2$ is closed.

Just a warning I have never studied topology. Thus, what I may have said is completely wrong. But I think I understand the question. - Jan 17th 2006, 07:56 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

does not work. What does is:

$\displaystyle (F_1\cup F_2)'=F_1'\cap F_2'$,

(where $\displaystyle A'$ denotes the complement of $\displaystyle A$)

but the intersection of two open sets (in this case $\displaystyle F_1'$ and $\displaystyle F_2'$), is open so

$\displaystyle (F_1\cup F_2)'$ is open, and so $\displaystyle F_1\cup F_2$ is closed.

.

RonL - Jan 18th 2006, 01:43 PMThePerfectHacker
Stupid me forgot to use de Morgan's Law's for sets :(

CaptainBlack is right.

But you (CaptainBlack) forgot to mention one important thing when you were destroying my proof. You forgot to state the the intersection of two sets is an element of the topological space. - Jan 18th 2006, 02:17 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

set (space) and O is the collection of subsets of X which are open.

X being a set inherits all the properties of sets, such as intersections and

unions of subsets of X also being subsets of X.

RonL - Jan 18th 2006, 02:22 PMThePerfectHacker
I see what you are saying. I just never studied topology, and did not know what $\displaystyle (X,O)$ stood for. When did you learn topology?

- Jan 18th 2006, 02:55 PMTexasGirlOne more...
For the same topological space, how do I show the following:

Let A be nonempty. If Fa contained in X is closed for each a contained in A, then ∩a Fa is closed. - Jan 18th 2006, 07:11 PMThePerfectHacker
This is how understand it $\displaystyle \bigcup_a F_a$ its complement is $\displaystyle \bigcap_a F_a$ ?

- Jan 18th 2006, 11:35 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

you should at least first look it up on MathWorld or on Wikipedia,

both of which are pretty good for this (I usualy do even if I am familiar with

the field).

When did I learn about topological spaces - as a final year undergraduate in

a course on Functional Analysis, about 30 years ago, but I don't see how

that is relevant to anything.

RonL - Jan 18th 2006, 11:45 PMCaptainBlackQuote:

Originally Posted by**TexasGirl**

Topological Space on Wikipedia for an definition of a topological space in terms

of closed sets, and de Morgan's Laws on PlanetMath for the stuff needed

to justify the Wikipedia statement.

But briefy (look at the references for justification):

$\displaystyle \left( \bigcup_{a\epsilon A} F_a' \right) '=\bigcap_{a\epsilon A} F_a$,

but the LHS is the complement of an open set and so closed so the RHS is

closed (an arbitary union of open sets is open, and all the $\displaystyle F_a$ are close

so all the $\displaystyle F_a'$ are open).

RonL - Jan 19th 2006, 12:42 PMThePerfectHacker
I always use wikipedia for math help. This is how I answered this topology question. I think wikipedia is the best link for math help.

- Jan 20th 2006, 10:07 AMThePerfectHacker
I was thinking about topological spaces and wondered if this has any importance.

1)Let $\displaystyle G$ be a group.

2)Let $\displaystyle G'=\{X\subseteq G\}$

3)Define $\displaystyle S=\{\}\cup G'$

4)Thus, $\displaystyle S$ is a topology on $\displaystyle G$

This is because the union/intersection of two groups is a itself a group and also because $\displaystyle S$ satisfies the topology axioms. - Jan 20th 2006, 10:42 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

binary operation defined on $\displaystyle G\times G$, such that the combination satisfies

the axioms for a group.

You have only used the set $\displaystyle G$ in your construction and none of

the properties of $\displaystyle \otimes$, so what you have done is in fact true

for any set.

Quote:

2)Let $\displaystyle G'=\{X\subseteq G\}$

$\displaystyle G'=\{X; \mbox{such that } X \subseteq G \}$

In which case this is the power set of $\displaystyle G$, usually written:

$\displaystyle \mathcal{P} (G)$.

Quote:

3)Define $\displaystyle S=\{\}\cup G'$

as $\displaystyle \emptyset \subset G$ and so an element of $\displaystyle \mathcal{P} (G)$ already (the null set is a subset of every set).

Quote:

4)Thus, $\displaystyle S$ is a topology on $\displaystyle G$

This is because the union/intersection of two groups is a itself a group and also because $\displaystyle S$ satisfies the topology axioms.

RonL - Jan 21st 2006, 06:04 PMThePerfectHacker
CaptainBlack what you said is true. I made a mistake in how I posed my question, I wanted to say a

**subgroup**not a**subset**.

The revised question:

1)Let $\displaystyle G$ be a group.

2)Let $\displaystyle P(G)=\{X|X\leq G\}$.

3)Let $\displaystyle S=\{\} \cup P(G)$.

4)Then $\displaystyle S$ is a topology on $\displaystyle G$.

The reason is because the union/intersection of two subgroups is a subgroup.