It is true that $\mathfrak{c} = \mathfrak{c}^{\aleph_0}$ but $\mathfrak{c} < \mathfrak{c}^\mathfrak{c}$.
- Is it true that $\mathfrak{c} = \mathfrak{c}^{X}$ for all $X < \mathfrak{c}$?
The answer is obviously trivial if we admit the CH. In fact, if the GCH holds we get:
$\aleph_n = \aleph_n^{X}$ for all $X < \aleph_n$ and $n$ a
However, even with the GCH, the property does not hold in general, since $\aleph_{\omega} < \aleph_{\omega}^{\aleph_0}$ (by König's theorem). So,
- For what cardinals $A$ does it hold that $A = A^B$ for all $A > B$?
- Is there a relation $\mathcal{R}$ such that $A = A^B$ for all $\mathcal{R} A B$?