Given two sets $A$ and $B$, is it true that $|B^A|=\max\{|B|,|\mathcal{P}(A)|\}$?

Question

My intuition is that this is true, because it makes sense for little sets (with cardinalities $|\mathbb{N}|,|2^\mathbb{N}|$). But i'm not sure at all that this is actually true.

If it's not, what's the real relation between $|B^A|$ and $|B|,|A|$.

(Obviously talking about infinite sets)

I've already proved that if $|B|\leq|A|$ then $|B^A|=|\mathcal{P}(A)|$. — R.V.N., Jan 30 '21 at 15:45
If the notation "B^A" ever seemed weird to you (as it did to me) it's worth speculating about why somebody may have chosen it. (It almost seems like a trick question, but it isn't.) — leslie townes, Jan 30 '21 at 15:51

mathcounterexamples.net · Answer 1 · 2021-01-30T16:05:27.860

1

Surely not true for finite sets. Counterexample

$$A=\{1,2\} \text{ and } B = \{1,2,3\}.$$

And if Singular cardinals hypothesis is assumed, the result is again wrong as

$$\text{if } \vert \mathcal P(A) \vert \lt \vert B \vert \text{ and } \text{cf} \vert B \vert \le \vert A \vert \text{ then } \vert B^A \vert = \vert B \vert^+$$

edited Jan 30 '21 at 16:05

answered Jan 30 '21 at 15:47

mathcounterexamples.net

70,018

Edited the post, just referring to infinite sets. – R.V.N. Jan 30 '21 at 15:59
Still wrong at least assuming SCH. – mathcounterexamples.net Jan 30 '21 at 16:06

Given two sets $A$ and $B$, is it true that $|B^A|=\max\{|B|,|\mathcal{P}(A)|\}$?

1 Answers1