Is it possible that |$\mathbb R$| is an inaccessible cardinal? I know that |$\mathbb R$| can be any cardinal with cofinaly not equal to $\aleph_0$. Does this include inaccessible cardinals?
And what would be the consequences if |$\mathbb R$| = $\kappa$?
 A: It depends what you mean by "inaccessible" - strongly inaccessible or weakly inaccessible.
Since a strongly inaccessible $\kappa$ is by definition uncountable and bigger than $2^\lambda$ for every $\lambda<\kappa$, we can't have $\vert\mathbb{R}\vert$ be strongly inaccessible.
On the other hand, we can have $\vert\mathbb{R}\vert$ be weakly inaccessible (that is, a regular limit cardinal), at least assuming that the existence of weakly inaccessible cardinals is consistent with ZFC. The proof that this is possible uses forcing, but if you've already seen forcing then here's the outline:


*

*Suppose $V$ is a model of ZFC with an inaccessible cardinal $\kappa$.

*Consider $L^V$, that is, $V$'s version of the constructible universe.

*In $L^V$, the Continuum Hypothesis holds and $\kappa$ is still weakly inaccessible - so in $L^V$ we have $2^{\aleph_0}<\kappa$.

*Now we force: add $\kappa$-many Cohen reals. Since this forcing is c.c.c., it preserves cardinals and cofinalities, so $\kappa$ remains a weakly inaccessible cardinal; but now $2^{\aleph_0}=\kappa$. (Actually, all that's obvious is that $2^{\aleph_0}\ge\kappa$; it takes an argument, using "nice names," to show that in fact we have equality. This isn't hard but it's worth pointing out.)
Meanwhile, as far as consequences of $2^{\aleph_0}$ being weakly inaccessible go, I'm not aware of any big ones beyond the obvious (e.g. CH isn't true).

Some further comments:


*

*In the proof sketch above, why did I need to pass to $L$ instead of working in $V$? Well, we could have in $V$ that $2^{\aleph_0}>\kappa$, in which case adding $\kappa$-many Cohen reals won't make $2^{\aleph_0}=\kappa$. Instead, if we want to handle this case without going to $L$ we need to collapse a cardinal: force with the set of partial maps from $\kappa$ to $\mathbb{R}$ whose domains have cardinality $<\kappa$. Since this is $(<\kappa)$-closed, $\kappa$ will stay weakly inaccessible.

*What if I try to run the same argument as above, but with $\kappa$ strongly inaccessible? Well, now in the last bullet point, I need to argue somehow that $\kappa$ remains strongly inaccessible after forcing, and this breaks down: specifically, there is no reason to have $2^\lambda<\kappa$ for all $\lambda<\kappa$ after we force.

*Finally, it's worth pointing out that there are senses in which $\kappa=\vert\mathbb{R}\vert$ can "appear" strongly inaccessible (or more): namely, by looking at $\kappa$ in an inner model $M$ (e.g. $M=L$). This inner model could be much smaller than $V$, and in particular it could be wrong about how the powerset function works: we'll always have $(2^\lambda)^M\le (2^\lambda)^V$, but not necessarily conversely. So it's totally possible to have e.g. "$2^{\aleph_0}$ is strongly inaccessible in $L$" (in fact, this will be true whenever $2^{\aleph_0}$ is weakly inaccessible, since weakly inaccessible cardinals are strongly inaccessible in $L$ since $L$ satisfies the GCH!). If you're interested in understanding how properties can differ between an inner model and the whole model, the term to look for is "absoluteness."
