# Topology Proof

1. Apr 21, 2010

### beetle2

1. The problem statement, all variables and given/known data

show that T:=(A subset X |A = 0 or X\A is finite) is a topology on X,
2. Relevant equations

We need to show 3 conditions.

1: X,0 are in T
2: The union of infinite open set are in T
3: The finite intersections of open sets are open.

3. The attempt at a solution

We see that $A \subset X$ is open in (T1 space) asX\A is finite

To show condition 1
if A = 0 the empty set it is in T
and A\X = X than it is in T.

To show 2

let $A \subset X$ open in T1 as X\A is finite

Let $\alpha \in I$ be an indexing set, $A_\alpha \in T$ so that $A \subset X$ be open as X\A is finite.

Than the $\cup_{\alpha \in I}$ X\$A_\alpha = \cap _{\alpha \in I}$ (X\$A_\alpha$)

Either each of the sets ( X\$A_\alpha$) = X , in which case the intersection is all of X, or at least one of them is finite , in which case the intersection is a subset of a finite set and hence finite.

To show 3

Let $A_1,A_2,A_3...A_n \subset X$be open as X\A is finite or all of X.

To show that $\cap A_{n} \in T$we must show that $\cap$ X\$A_n$ is either finite or all of X.

But $\cap X$\$A_{n} = \cup X$\$A_{n}$.

Either this set is a union of finite sets and hence finite, or for some X\$A_{i} i \in I = X$and the union is all of X.

Thus (A,T) is a topological space.

2. Apr 21, 2010

### VeeEight

You have the right idea for all of the conditions. Applying the set theory properties was the key.

Have you heard of the countable complement topology?