[증명] X is Hausdorff if only if the diagonal ∆ = { x × x | x ∈ X } is closed in X × X.

명제.

X is Hausdorff if only if the diagonal ∆ = { x × x | x ∈ X } is closed in X × X.

 

 

증명.

 

X is hausdorff 

⇒ ∀x ≠ y ∈ X, ∃Nx, Ny s.t. Nx ∩ Ny = ∅

⇒ Nx × Ny is open in X × X, Nx × Ny ∩ ∆ = ∅

 

why?

if not, ∃ w × z ∈ X × X s.t.  w ∈ Nx, z ∈ Ny, w × z ∈ ∆.

from w ∈ Nx, z ∈ Ny, we know w ≠ z because Nx ∩ Ny = ∅

from w × z ∈ ∆, we know w = z

contradiction exists.

 

Now, consider D = X × X - ∆, then D = { x × y | x ≠ y ∈ X }

 

⇒ ∀ x × y ∈ D, ∃ open set Nx × Ny  s.t. x × y ⊂ Nx × Ny ⊂ D

⇒ D is open

⇒ ∆ is closed

 

Now, start with that ∆ is closed

⇒ D is open

⇒ ∀ x × y ∈ D, ∃ open set C  s.t. x × y ⊂ C ⊂ D

⇒ ∃ open set Nx, Ny ⊂ X s.t. x ∈ Nx and y ∈ Ny and  x × y ⊂ Nx × Ny ⊂ C ⊂ D

⇒ Nx × Ny ⊂ D

⇒ ∀ w ∈ Nx and z ∈ Ny, w ≠ z

⇒ Nx ∩ Ny = ∅

⇒ X is hausdorff 

 

결론

X is Hausdorff ⇔ the diagonal ∆ = { x × x | x ∈ X } is closed in X × X.

 

COROLLARY

 

In the purple-colored part in the beginning of the proof,

 ∀x ≠ y ∈ X, 

substitute ≠ with <

then we can also get the conclusion that

 W = { x × y | x ≥ y ∈ X } is closed in X × X.

and with >, vice versa. (IF given it is ordered)