Is the Intersection of Closed Sets in a Topological Space Also Closed?

[son]

Prove that the intersection of any collection of closed sets in a
topological space X is closed.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[SammyS]

What have you tried?

Where are you stuck?

How is a closed set defined?

 

[lanedance]

think about complements

 

[culturedmath]

I believe, the shortest and the easiest way is to start with the definition of a closed set and what an intersection is.

Good Luck.

 

[HallsofIvy]

Indeed, the shortest and easiest way to prove anything is to start with the definitions!

However, there are many different ways to define "closed set". son, what definition are you using?

 

[son]

the theorem i am using for a closed set is...
Let X be a topological space. the following statements about the collection of closed set in X hold:
(i) the empty set and X are closed
(ii) the intersection of any collection of closed sets is a closed set
(iii) the union of finitely many closed sets is a closed set

 

[culturedmath]

I am new in the forum ( although I have read it for some time ) and I am not quite sure how much of a hint I am allowed to give you but:

You can prove that a set is closed using "balls". I would suggest you to work in this direction.

 

[micromass]

the theorem i am using for a closed set is...
Let X be a topological space. the following statements about the collection of closed set in X hold:
(i) the empty set and X are closed
(ii) the intersection of any collection of closed sets is a closed set
(iii) the union of finitely many closed sets is a closed set

Yes, that's the theorem you are trying to prove. But what is the definition of a closed set??

 

[Landau]

This follows directly from the definition. Please show us what part you're having trouble with. This is not a homework answer generator.

 

[SammyS]

A closed set is the complement of ___?___ .

 

[son]

the definition of a closed set is... a subset A of a topological space X is closed if the set X - A is open.

but I am not sure how i would start the proof...

 

[son]

the definition of a closed set is... a subset A of a topological space X is closed if the set X - A is open.

but I am not sure how i would start the proof...

 

[jbunniii]

What is the definition of a topological space? Doesn't it say something about the union of a collection of open sets?

 

[SammyS]

the definition of a closed set is... a subset A of a topological space X is closed if the set X - A is open.

but I'm not sure how i would start the proof...

I would start it something like:

\text{Let }\left\{A_\alpha\right\}\text{ be an arbitrary collection of closed sets in a topological space }X, \text{ where }\alpha\in I,\ I \text{ an indexing set.}

\text{The set }C=\bigcup_{\alpha\in I}\,A_\alpha\ \text{ is the union of an arbitrary collection of closed sets in }X.

...

Now show that the compliment of set C is open in X.

 

[son]

Let F={F_i} be a collection of closed sets. Then F_i=X-U_i for some collection {U_i} of open sets of X, because of the definition of closed. Then De-Morgans rules give
intersection F_i = intersection (X-U_i) = X - (union U_i)
But union U_i is an open set because the unions of open sets are open. Thus the set on the right hand side of the above equation is the complement of an open set. Hence the intersection F_i is closed.

does this look RIGHT??

 

[SammyS]

That's the general idea. It could be a bit more polished.

 

[HallsofIvy]

I am new in the forum ( although I have read it for some time ) and I am not quite sure how much of a hint I am allowed to give you but:

You can prove that a set is closed using "balls". I would suggest you to work in this direction.

No, you cannot. That works only in a "metric space" because balls are only defined in a metric space. This problem clearly is about general topological spaces.

 

[Fredrik]

Suppose that \{F_i|i\in I\} is a collection of closed sets. You want to prove that

\bigcap_{i\in I}F_i

is closed. By the definition you posted, this is the same thing as showing that

\Big(\bigcap_{i\in I}F_i\Big)^c

is open. Can you think of a way to rewrite this last expression as something that's obviously open?