# Exponential of cardinal numbers

there is two wrong statement that I want to find counterexample for them. if $\alpha$ and $\beta$ and $\gamma$ be infinite cardinals then show that these two statements are wrong

• $\alpha < \beta \Longrightarrow \alpha^\gamma < \beta^\gamma$

for this one if $\alpha = \aleph_0$ and $\beta = \gamma = \aleph_1$ then we have $\aleph_1^{\aleph_1} = (2^{\aleph_0})^{\aleph_1} = 2^{\aleph_0 \aleph_1} = 2^{\aleph_1}$ and because $2^{\aleph_1} \leq \aleph_0^{\aleph_1} \leq \aleph_1^{\aleph_1}$ we have that $\aleph_0^{\aleph_1} = \aleph_1^{\aleph_1}$. is this right?

and the second one is

• $\alpha < \beta \Longrightarrow 2^\alpha < 2^\beta$

for this one if $card(A) = \alpha$ and $card(B) = \beta$ we have that $2^\alpha = card(P(A))$ and $2^\beta = card(P(B))$ and I think this statement is true when $\alpha < \beta$ then $card(P(A)) < card(P(B))$ is this right?

it's from Iran national exam of Computer Science of two years ago.

• The second one looks tempting, but I'd suspect that it depends. – Hagen von Eitzen Apr 3 '16 at 11:12

(I'll assume choice throughout.)

You're assuming the continuum hypothesis in saying that $\aleph_1 = 2^{\aleph_0}$; I don't know if you're allowed to do that.

If $\gamma > 2^{\aleph_0}$ then we can use $$2^{\gamma} \leq \aleph_0^{\gamma} \leq (2^{\aleph_0})^{\gamma} = 2^{\aleph_0 \gamma} = 2^{\gamma}$$ so equality holds throughout, and in particular $$\aleph_0^{\gamma} = (2^{\aleph_0})^{\gamma}$$ even though $\aleph_0 < 2^{\aleph_0}$.

For the second one: I don't believe your argument, I'm afraid, because it's possible to force $2^{\aleph_0} = 2^{\aleph_1}$, so it's relatively consistent that there is a counterexample.

However, the Generalised Continuum Hypothesis says that $2^{\alpha} = \alpha^+$, so under GCH we would have that $\alpha < \beta \Rightarrow \alpha^+ < \beta^+ \Rightarrow 2^{\alpha} < 2^{\beta}$. It's (relatively-to-ZFC) consistent that GCH holds, so it's relatively consistent that there are no counterexamples.

Notice that any counterexample for the second one will give you a counterexample for the first one, since $2^{\beta} \leq \beta^{\beta} \leq 2^{\beta \times \beta} = 2^{\beta}$, so equality holds throughout.

Therefore letting $\beta = \gamma$, if we have $\alpha < \beta$ such that $2^{\alpha} \geq 2^{\beta}$, then we have $\alpha < \beta$ such that $2^{\alpha} \geq \beta^{\beta} = \beta^{\gamma}$.