Which powers of the closed unit interval are homeomorphic?

Question

It is known that no two distinct finite powers of the closed unit interval are homeomorphic:

$I^m$ is homeomorphic to $I^n$ iff $m=n$. (Brouwer, Lebesgue, 1911)

Is the analogous result for infinite powers of $I$ true?

That is, is it true that $I^\alpha$ is homeomorphic to $I^\beta$ iff $\alpha=\beta$, for cardinal numbers $\alpha$ and $\beta$?

(An affirmative answer would give a hope of defining a transfinite inductive dimension for all spaces, a problem essentially posed by Carl Menger.)

If not, what are the homeomorphism classes among the infinite powers of $I$?

Added Later: Since $|I^\alpha|=2^{\aleph _0 \alpha}=2^\alpha$ whenever $\alpha$ is infinite, assuming the Generalised Continuum Hypothesis, of course, $2^\alpha=2^\beta \Rightarrow \alpha=\beta$ and $I^\alpha$ and $I^\beta$ have different cardinalities if $\alpha$ and $\beta$ are distinct. But the answer below shows that they can be proved to be not homeomorphic within ZFC, and, more interestingly, even when they have the same cardinality (in some model which violates GCH).

Answer 1

Yes. $I^{\alpha}=I^{\beta}$ does imply that $\alpha=\beta$. To see this, suppose that $\alpha$ is an infinite cardinal. Then each point in $I^{\alpha}$ is the intersection of $\alpha$ many open sets, but each point in $I^{\alpha}$ is not the intersection of less than $\alpha$ many open sets.

In greater detail, if $(x_{\beta})_{\beta<\alpha}\in I^{\alpha}$, then for each $n\in\mathbb{N}$ and $J\subseteq\alpha$ with $|J|<\aleph_{0}$, let $U_{J,n}=\{(y_{\beta})_{\beta<\alpha}:|y_{j}-x_{j}|<\frac{1}{n}\,\textrm{for}\,j\in J\}$. Then $(x_{\beta})_{\beta<\alpha}=\bigcap_{J,n}U_{J,n}$, so $(x_{\beta})_{\beta<\alpha}$ is the intersection of $\alpha$ many open sets.

Now, suppose $|J|<\alpha$ and $U_{i}\subseteq I^{\alpha}$ is an open neighborhood of $(x_{\beta})_{\beta<\alpha}$ for $i\in J$. Then for each $i\in J$ there is a finite $J_{i}\subseteq\alpha$ where if $(y_{\beta})_{\beta\in\alpha}\in I^{\alpha}$ and $y_{j}=x_{j}$ for $j\in J_{i}$, then $(y_{\beta})_{\beta\in\alpha}\in U_{i}$. Since $|\bigcup_{i\in J}J_{i}|=|J|<\alpha$, $\bigcup_{i\in J}J_{i}\neq\alpha$. Therefore, whenever $y_{\beta}=x_{\beta}$ for $\beta\in\bigcup_{i\in J}J_{i}$, we have $(y_{\beta})_{\beta\in \alpha}\in\bigcap_{i\in J}U_{i}$, but we do not necessarily have $(y_{\beta})_{\beta\in \alpha}=(x_{\beta})_{\beta<\alpha}$. Hence, $\bigcap_{i\in J}U_{i}$ is not equal to $\{(x_{\beta})_{\beta<\alpha}\}$.