Does $|\Bbb R^I| = |\Bbb R|$ imply that $I$ is (at most) countable?

Question

$\newcommand{\R}{\Bbb R}$ This question is inspired from this question where the OP was inquiring about the cardinality of $\Bbb R^I$ for different $I$.

As in my answer, for $I$ finite or countable, we have that $|\R^I| = |\R|$. For $I = \R$, we have $|\R^I| = |2^\R| > |\R|.$

However, for $I$ with cardinality strictly between $|\Bbb N|$ and $|\R|$, can we say anything?

I am aware that $\sf{CH}$ states that there is no such set. For now, let us assume $\sf{AC}$ or at least the hypothesis that any two cardinalities are comparable. Does $|\R^I| = |\R|$ then imply that $|I| \le |\Bbb N|$?

My work so far has shown that $|\R^I| = |R| \implies |I| < |\R|$.

Answer 1

Suppose $|I| = \aleph_1$ where $\aleph_0 < \aleph_1 < \mathfrak{c}=2^{\aleph_0}$, which can hold in some models of set theory.

Then $2^{|I|}$ is at least $2^{\aleph_0}$, but could be more or equal, depending on what model of set theory you are (e.g. MA (Martin's axiom) will imply they're equal). See Easton's theorem for other options.

So in short: you cannot even be sure there will be intermediate cardinalities, but if there are we cannot say more than that it's at least $\mathfrak{c}$. So the short answer to the title question is no it need not be.

Answer 2

Even under $\mathrm{GCH}$, if $\lambda$ is a cardinal with $\text{cf } \mathfrak c \le \lambda \lt \mathfrak c$, you have $\mathfrak c^\lambda = \mathfrak c^+$.

See Jech - Set Theory, theorem 5.15.