# Topology question

#### Jomo

##### Elite Member
I was told that something is wrong in the proof below.
Any definition that I did not know I looked up.
I however do not see any errors. It's not the best written proof.

Theorem 1: Suppose X is an infinite set and
T is the finite-closed topology on X. If S is a subset X, then S is open iff
S is infinite or S is empty.

Proof: (=>)Suppose S is open in the finite-closed topology. If S is empty
we are done, so assume S is not empty. Since S is open, S is not closed, which
(by DeMorgan's law) means that S /= X and S is not finite. Thus S
is infinite.
(<=) Suppose S is either infinite or empty. If S is empty, then it is
open by denition of a topology. If S is infinite, then it is not finite so
it is not closed. Therefore S is open in the finite-closed topology.

#### Jomo

##### Elite Member
I think the theorem is false. Here is counter example. Let X be integers and S be even integers. S is infinite but not open. One mistake in proof I just realized is that a set can be both open and closed.

#### pka

##### Elite Member
Theorem 1: Suppose X is an infinite set and
T is the finite-closed topology on X. If S is a subset X, then S is open iff
S is infinite or S is empty.
What is the definition of the finite-closed topology?
I know what the finite-complement topology on a set is.
And the above theorem works in such a top-space.