[증명] Simple ordered set is Hausdorff

Theorem 명제. Every simple ordered set in the order topology is Hausdorff

 

Proof 증명

 

Let X be a simple ordered set in the order topology

 

∀x<y∈X,

case 1. ∃z∈X s.t. x<z<y

there are neighborhoods Nx, Ny⊂X s.t.

Nx= (a,z) (if ∃a<x) or [x,z) (if x is smallest element)

Ny= (z,b) (if ∃b>y) or (z,y] (if y is largest element)

 

then Nx∩Ny=∅

 

case 2. such z doesn't exist

there are neighborhoods Nx, Ny⊂X s.t.

Nx= (a,y) (if ∃a<x) or [x,y) (if x is smallest element)

Ny= (x,b) (if ∃b>y) or (x,y] (if y is largest element)

 

Now, consider Nx∩Ny={z|x<z<y}

such z doesn't exist, so Nx∩Ny=∅

 

 

Either case, there exists neighborhoods Nx, Ny⊂X s.t. Nx∩Ny=∅

 

Therefore, X is hausdorff