Is the Countable Complement Topology a Valid Topological Space?

[beetle2]

Homework Statement

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

Homework 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.

The Attempt at a Solution

We see that [itex]A \subset X [/itex] 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 [itex]A \subset X [/itex] open in T1 as X\A is finite

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

Than the [itex]\cup_{\alpha \in I} [/itex] X\[itex]A_\alpha = \cap _{\alpha \in I}[/itex] (X\[itex]A_\alpha[/itex])

Either each of the sets ( X\[itex]A_\alpha[/itex]) = 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 [itex]A_1,A_2,A_3...A_n \subset X [/itex]be open as X\A is finite or all of X.

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

But [itex]\cap X[/itex]\[itex]A_{n} = \cup X[/itex]\[itex]A_{n}[/itex].

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


Thus (A,T) is a topological space.

 

[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?