No Hereditarily Normalilty In Uncountable Product

In a previous post, I showed that the uncountable product of the space of the integers is not normal. Consequently, the uncountable product of the unit interval is never hereditarily normal since it contains a copy of the uncountable product of the integers. It turns out that, as the title of this post suggests, hereditarily normality cannot happen in uncountable product of spaces. Specifically the product of uncountably many spaces, each of which has at least two points, is never hereditarily normal.

Let Y=\Pi_{\alpha \in A} X_\alpha such that \vert A \vert \geq \omega_1 and each X_\alpha has at least two points. The proof is quite simple, a matter of rearranging the factors so that the product space Y contains a copy of \Pi_{\alpha \in A} \mathbb{N}.

For each \alpha \in A, let Z_\alpha \subset X_\alpha be a two-point subspace. Break up A into \vert A \vert many disjoint subsets A_z, z \in A where \vert A_z \vert=\omega. For each z \in A, let C_z=\Pi_{h \in A_z} Z_h, which is homeomorphic to the middle third Cantor set in the unit interval. Thus, \Pi_{\alpha \in A}Z_\alpha=\Pi_{z \in A}C_z \subset \Pi_{\alpha \in A}X_\alpha.

Each Cantor set contains a subset that is homeomorphic to the discrete space of the natural numbers \mathbb{N}. Thus \Pi_{z \in A}\mathbb{N} \subset \Pi_{z \in A}C_z.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s