Prove every Hausdorff topology on a finite set is discret.

[Hodgey8806]

Homework Statement

Prove that every Hausdorff topology on a finite set is discrete.
I'm trying to understand a proof of this, but it's throwing me off--here's why:

Homework Equations

To be Hausdorff means for any two distinct points, there exists disjoint neighborhoods for those points.
Also, any finite subset of a Hausdorff space is closed.

The Attempt at a Solution

Let a set X have n elements (I'll write it more formal later), but I'll denote them a 1,...,i,...,j,...n.
For each singleton element, we can write write it as:
{i} = $\bigcap$(X\{j}) s.t. j≠i.
And the set {i} is open because it's the intersection of open sets (X\{j}).

However, isn't that opposite of Hausdorff because both sets are finite subsets.

Thank you in advance for your help.

 

[Dick]

{i} is closed, since it's a finite subset of a Hausdorff space. Now you've proved {i} is also open. So it's clopen. Doesn't that make it a discrete space?

 

[Hodgey8806]

Ah ok!

My book mentions that every subset of a discrete space is closed, but it doesn't explictly say that it is open when we first discussed them. It mentioned the topology is that every set is open...so I suppose it's implied.

Thank you!

 

[Dick]

Ah ok!

My book mentions that every subset of a discrete space is closed, but it doesn't explictly say that it is open when we first discussed them. It mentioned the topology is that every set is open...so I suppose it's implied.

Thank you!

Right. If every set S is closed then its complement is also closed. So S is also open.