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

[Hodgey8806]

Homework Statement

Prove that T₁={U subset of X: X\U is finite or is all of X} is a topology.

Homework Equations

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

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 {U_a:a is in A} be a collection of sets in T₁
Spse U_a ≠ X.
X\(indexed unionU_a) = indexed intersection(X\U_a)
This is finite since each X\U_a is finite or is all of X. (element of T₁)

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

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

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

 

[Bacle2]

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

 

[HallsofIvy]

Homework Statement

Prove that T₁={U subset of X: X\U is finite or is all of X} is a topology.

Homework Equations

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

The Attempt at a Solution

1) X/X = empty set, finite. Thus X is in T₁
2) X/empty set = X, infinite.

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

Thus Empty Set is in T₁
3) Let {U_a:a is in A} be a collection of sets in T₁
Spse U_a ≠ X.
X\(indexed unionU_a) = indexed intersection(X\U_a)
This is finite since each X\U_a is finite or is all of X. (element of T₁)

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

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

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

 

[SammyS]

Hodgey,

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

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

For (3.):

Let \displaystyle <br /> \left\{ \text{U}_\alpha\ :\ \alpha \in \text{A}\right\} be an arbitrary collection of sets in T₁ .
[STRIKE]Spse U_a ≠ 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 <br /> \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.​

 

[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!