# Cardinal exponentiation

My understanding of exponentiation of cardinals leads to the conclusion that if $$2 \leq \kappa \leq \lambda$$, then $$2^\lambda = \kappa^\lambda$$ , because:

$$2^\lambda \leq \kappa^\lambda \leq (2^\kappa)^\lambda = 2^{\kappa \cdot \lambda} = 2^\lambda$$

However, I am slightly confused at this because Cantor’s Theorem states that $$\lvert A \rvert < \lvert \mathcal{P}(A) \rvert$$, for every set A. Hence, $$\kappa < 2^\kappa$$, for every cardinal $$\kappa$$

I know it would lead to the inconsistent conclusion that $$2^\lambda < 2^\lambda$$ but why doesn't this strict inequality apply in the second step of the equation above?

• $\kappa\mapsto2^\kappa$ is not preserving strict order. So there is no contradiction. – Asaf Karagila May 22 '19 at 0:13
• This question is more appropriate for math.stackexchange, but as a quick comment: re: "why doesn't this strict inequality apply in the second step of the equation above?," why should it? – Noah Schweber May 22 '19 at 1:34
• "why should it?" - We have $\kappa < 2^\kappa$ by Cantor's inequality (which holds even for infinite cardinals), but this becomes $\kappa^\lambda \leq (2^\kappa)^\lambda$ after exponentiation. So what is it about the exponentiation that allows equality to creep in here when it started as a strict inequality? – Corey May 22 '19 at 2:43
• Perhaps my missing understanding was simpler than I thought, as even addition and multiplication operations (on infinite cardinals) can create a similar situation. In effect, it may be akin to how "multiply by negative" flips inequalities, and "divide by zero" just outright breaks (in)equalities. "Infinite cardinal operations" can potentially introduce equality (due, I assume as Asaf responded, to these operations not preserving strict order) Thank you all. – Corey May 22 '19 at 3:42
• In fact, in general it is difficult to obtain strict inequalities in cardinal arithmetic when doing operations; that is, if $\lambda\lt \kappa$, and you do an operation to both sides, in general it is difficult to preserve the strict inequality: adding, multiplying, taking power sets, none of those guarantee a strict inequality. Koenig’s Theorem is one of the few results in ZFC that yields a strict inequality. – Arturo Magidin May 22 '19 at 4:18

• Cantor's theorem compares base to power: it might help to rewrite it as $$\kappa^2<2^\kappa.$$
• The fact that $$\rho^\kappa$$ can equal $$\lambda^\kappa$$ even if $$\rho<\lambda$$ compares base to base.
• Cantor's Theorem extends to infinite sets too. (Basic proof involves assuming there is a function $f:A \to \mathcal{P}(A)$ that is surjective, but then considering the set $S =\{a \in A \mid a \notin f(x)\}$ which leads to a contradiction). But yes, I believe it is the infinite exponent of $\lambda$ that is causing the strict inequality of $\kappa < 2^\kappa$ to become equality in $\kappa^\lambda = 2^\lambda$. The same can be seen to occur under infinite addition ($\kappa + \lambda = 2 + \lambda$) and infinite multiplication ($\kappa \cdot \lambda = 2 \cdot \lambda$). Thanks – Corey May 23 '19 at 1:51
Since $$\kappa\lt2^\kappa$$, $$\kappa^\lambda\le2^{\kappa\lambda}=2^\lambda$$. You've proven this second inequality isn't strict. There's only an inconsistency if you can prove $$\kappa\lt2^\kappa$$ would imply $$\kappa^\lambda\lt2^{\kappa\lambda}$$ as well. If you can prove this, let us know; it'd be exciting to know ZFC is inconsistent. In the mean time, we have a proof that $$\kappa^\lambda\not\lt2^{\kappa\lambda}$$, so let's just accept that.