Paracompact Hausdorff implies binormal

Statement
Any paracompact Hausdorff space (i.e., a space that is both paracompact and Hausdorff) is a binormal space.

Paracompact Hausdorff space
A topological space $$X$$ is termed a paracompact Hausdorff space if it satisfies the following two conditions:


 * 1) It is a paracompact space: every open cover has a locally finite open refinement.
 * 2) It is a Hausdorff space: any two points can be separated by disjoint open subsets.

Binormal space
A topological space $$X$$ is termed a binormal space if the product of $$X$$ and the unit interval $$[0,1]$$ (equipped with the Euclidean topology) is a normal space.

Facts used

 * 1) uses::Compact times paracompact implies paracompact
 * 2) uses::Hausdorffness is product-closed
 * 3) uses::Paracompact Hausdorff implies normal

Proof
Given: A paracompact Hausdorff space $$X$$. $$I = [0,1]$$ is the unit interval.

To prove: The space $$X \times I$$ is a normal space.

Proof:


 * 1) $$X \times I$$ is paracompact: This follows from $$X$$ being paracompact, $$I$$ being compact, and fact (1).
 * 2) $$X \times I$$ is Hausdorff: This follows from fact (2).
 * 3) $$X \times I$$ is normal: The previous two steps yield that $$X \times I$$ is paracompact Hausdorff. Fact (3) now yields that $$X \times I$$ is normal.