#### Cratylus

##### New member

- Joined
- Aug 14, 2020

- Messages
- 32

**4.1 Definition**Let X. be a topological space and let x\(\displaystyle \in\) X . A set N\(\displaystyle \in\) X is a

neighborhood of x if there an open set U\(\displaystyle \in\) X s.t x\(\displaystyle \in\) U\(\displaystyle \subset\) 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 \(\displaystyle \ne\)y, there are two open sets U and V, s.t x\(\displaystyle \in\) U ,y\(\displaystyle \in\) V

with U \(\displaystyle \cap\) V=0. But X\S is open and x\(\displaystyle \in\) U\(\displaystyle \subset\) N\S

and y\(\displaystyle \in\)V\(\displaystyle \subset\) N\S in S. with U \(\displaystyle \cap\) 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.