#### Cratylus

##### New member

- Joined
- Aug 14, 2020

- Messages
- 40

**4.1 Definition**Let X. be a topological space and let x[MATH]\in[/MATH] X . A set N[MATH]\in[/MATH] X is a

neighborhood of x if there an open set U[MATH]\in[/MATH] X s.t x[MATH]\in[/MATH] U[MATH]\subset[/MATH] N

**Attempted Proof(via contradiction)**

Let (X,T) be a topology.S being the set of countable ordinals, it is Hausdorff.(They many open sets contained in

S)

This implies that if x [MATH]\ne[/MATH]y, there are two open sets U and V, s.t x[MATH]\in[/MATH] U ,y[MATH]\in[/MATH] V

with U [MATH]\cap[/MATH] V=0. But X\S is open and x[MATH]\in[/MATH] U[MATH]\subset[/MATH] N\S

and y[MATH]\in[/MATH]V[MATH]\subset[/MATH] N\S in S. with U [MATH]\cap[/MATH] V=0

But this is impossible . Thus S is not closed in X

I am using the book A First Course in Topology by Robert Conover

l can assume everythinh up to this point and info on ordinals.

Any help would be appreciated.