# The Uncountable Product Of The Unit Interval Is Not Hereditarily Separable

Let $I=[0,1]$. The product space $I^I$ is separable, as is the product space $I^{\omega_1}$. By a theorem of [Ross & Stone], the product of $\leq$ continuum many separable spaces is a separable space. However, both $I^I$ and $I^{\omega_1}$ are not hereditarily separable. In a previous post on the first uncountable ordinal, I showed that the space of continuous functions on $[0,\omega_1)$ is not separable (see the post). I show in this post that the product space $I^{\omega_1}$ contains a copy of this function space. Hence both $I^I$ and $I^{\omega_1}$ are not hereditarily separable.

In my previous post, I showed that $C_p([0,\omega_1))$ is not separable. Note that $\mathbb{R}^{\omega_1}$ contains $C_p([0,\omega_1))$ as a subspace. Thus $\mathbb{R}^{\omega_1}$ is not hereditarily separable. Note that $\mathbb{R}^{\omega_1}$ is considered a subspace of $I^{\omega_1}$. Thus $I^{\omega_1}$ is not hereditarily separable. Note that $I^{\omega_1}$ is considered a subspace of $I^I$. If the cardinality of $I$ is greater than $\omega_1$, then all functions in $I^I$ that are zero on all points of $I-\omega_1$ would be a copy of $I^{\omega_1}$. If the cardinality of $I$ equals $\omega_1$, $I^{\omega_1}=I^I$Thus both $I^I$ and $I^{\omega_1}$ are not hereditarily separable.

Reference
[Ross & Stone]
Ross, K. A. and Stone, A. H. [1964] Products of Separable Spaces, The American Mathematical Monthly, Vol 71, No. 4, pp. 398-403.