Let be the discrete space the natural numbers. It is known that is not normal (see a proof here). It turns out that the product of uncountably many non-compact metric spaces is never normal. In 1948, A. H. Stone proved that the uncountable product of metric spaces is normal if and only if all but countably many factors are compact. Thus product of uncountably many copies of is never normal.

**Theorem**. Let be a family of metrizable spaces. The following conditions are equivalent.

(1) is paracompact.

(2) is normal.

(3) All but countably many are compact.

**Proof**. is obvious.

. Suppose is normal. I Claim that all but countably many factors are countably compact. When this claim is established, (3) is established. Note that if a paracompact space is countably compact, it is compact. Suppose that there are uncountably many that are not countably compact where . Then each such would contain a closed copy of . Thus the product space would contains as a closed subspace. This is a contradiction since any closed subspace of a normal space is normal. Thus all but countably many factors are countably compact.

. Suppose all but countably many are compact. Then where is the product of all the compact factors and is the product of the countably many non-compact factors. Note that is also a metrizable space. The product of a compact space and a paracompact space is paracompact (see a proof here). Thus (1) is established.

**Corollary**. Let be a family of separable metrizable spaces. The following conditions are equivalent.

(1) is paracompact.

(2) is Lindelof.

(3) is normal.

(4) All but countably many are compact.

The only thing I want to mention about the corollary is that being a product of separable spaces, the product space has the countable chain condition (ccc). In any space with the ccc, paracompactness implies the Lindelof property (see a proof here).

**Reference**

[Stone] Stone, A. H., [1948] *Paracompact and Product Spaces*, Bull. Amer. Math. Soc., 54, 977-982.

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