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^IThus 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.

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