Prove that the following is a Topology. I really just want to clean it up.

1. Sep 1, 2012

Hodgey8806

1. The problem statement, all variables and given/known data
Prove that T1={U subset of X: X\U is finite or is all of X} is a topology.

2. Relevant equations
DeMorgan's Laws will be useful.
Empty set is defined as finite, and X is an arbitrary infinite set.

3. The attempt at a solution
1) X/X = empty set, finite. Thus X is in T1
2) X/empty set = X, infinite. Thus Empty Set is in T1
3) Let {Ua:a is in A} be a collection of sets in T1
Spse Ua ≠ X.
X\(indexed unionUa) = indexed intersection(X\Ua)
This is finite since each X\Ua is finite or is all of X. (element of T1)

Spse Ua = empty set for all a, then the indexed intersection(X\Ua) = X (element of T1)
Spse Ua-bar ≠ empty set for some a, then the indexed intersection(X\Ua) is a subset of X\Ua-bar (element of T1).

4) Let {Ua:a is in A} be a collection of sets in T1
Spse Ua ≠ empty set.
X\(indexed intersectionUa) = indexed union(X\Ua),
Since the number of intersection must be finite and X\Ua is finite for all a,
Then the union of finitely many finite sets is finite. (element of T1)

If Ua = empty set for some a,
Then the indexed union (X\Ua) is either finite or all of X. (element of T1)

2. Sep 2, 2012

Bacle2

I,m sorry, Idon't get what you're asking.

3. Sep 2, 2012

HallsofIvy

Staff Emeritus
Actually, you are NOT given that X is infinite. But the point is that X/empty set is all of X.

Last edited by a moderator: Sep 2, 2012
4. Sep 2, 2012

SammyS

Staff Emeritus
Hodgey,

(1.) & (2.) look fine.

I'll edit & make some comments below for the others.

For (3.):
Let $\displaystyle \left\{ \text{U}_\alpha\ :\ \alpha \in \text{A}\right\}$ be an arbitrary collection of sets in T1 .
[STRIKE]Spse Ua ≠ X. [/STRIKE] (Not needed)

$\displaystyle \text{X}\backslash \left(\bigcup\left(\text{U}_\alpha\right)\right)= \bigcap\left(\text{X}\backslash\text{U}_\alpha \right)$
If $\displaystyle \text{U}_\alpha = \emptyset \text{ for all }\alpha\in\text{A}\,,$ then $\displaystyle \bigcap\left(\text{X}\backslash\text{U}_\alpha \right)=\text{X}\ .$

Otherwise, argue that one of the $\displaystyle \text{X}\backslash\text{U}_\alpha$ is finite for some $\displaystyle \text{U}_\alpha$ ... so that the intersection is finite.
...

For (4):
This should be a finite intersection, so a finite collection of sets.​

5. Sep 2, 2012

Hodgey8806

Thank you very much! I understand this a bit better now. To me that piece didn't seem necessary, but we worked through it fast in class so I suppose it was there for good measure.

I argued on 3) that it is finite because if it intersects a finite set, the indexed intersection is now a subset of that particular finite set.

Thanks again! I have many more questions I need guidance on!