# 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 $\Pi_{z \in A}\mathbb{N} \subset \Pi_{z \in A}C_z$.

Advertisements