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I'm having a hard time getting over the following question, which appears in Schimmerling's "A Course on Set Theory."

(Problem) Given that $\kappa$ and $\lambda$ are infinite cardinals with $\kappa\le\lambda$, show that

$$|\{X\subseteq\lambda:|X|=\kappa\}|=\lambda^\kappa.$$

Denoting the set on the left hand side of the equation by $S$, it's not difficult to see that $S\subseteq\lambda^\kappa$. Indeed, $S$ is the set of all bijective mappings from $\kappa$ to some subset of $\lambda$ -- in other words, injections from $\kappa$ to $\lambda$. But $\lambda^\kappa$ also contains non-injective mappings from $\kappa$ to $\lambda$, and so it appears absurd that the cardinality of $S$, that is the least ordinal that could be bijected to $S$, could be its superset $\lambda^\kappa$.

Regardless, I'm trying to find a nice injection from $\lambda^\kappa$ to $S$ so that I could apply Cantor-Schroeder-Bernstein. But I haven't been able to find one so far.

Could someone explain the problem with my intuition, or provide some hints as to thinking about the question? Thanks!

youngtableaux
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This really isn't a good question to ask, since (as pointed out by Asaf) it has already been answered. (See comments above.) I should feel better by trying to explain the answer and resolve the issue of absurdity.

Mirroring the explanation provided here, we could view every function $f:\kappa\rightarrow\lambda$ by its graph, giving rise to pairs of elements $(\alpha,\beta)\in\kappa\times\lambda$. It doesn't matter whether $f$ is injective or not; the set $S_f=\{(\alpha,\beta):f(\alpha)=\beta\}$ contains $\kappa$ elements. Note also that $S_f\subseteq \kappa\times\lambda$. By the rules of cardinal multiplication $\kappa\times\lambda=\max({\lambda,\kappa})=\lambda$, and so there exists some subset of $\lambda$ that corresponds to $S_f$ in a bijection between $\kappa\times\lambda$ and $\lambda$.

All of this implies that for every function $f$ from $\kappa$ to $\lambda$, that is $f\in\lambda^\kappa$, there is a set $X_f\subseteq\lambda$ that corresponds to a set of size (i.e., cardinality) $\kappa$. By Cantor-Schroeder-Bernstein, which says that if for two sets $A$ and $B$ there exist an injection from $A$ to $B$ and another from $B$ to $A$ then $A$ can be bijected to $B$ (and so $|A|=|B|$), the equation holds.

Looking back, the trick is to think of functions as pairs of elements, or to provide some concrete description of it (at least more than just thinking about the properties of injectivity). This trick appears in, for instance, the question of showing that $(0,1)\sim[0,1)$. In one of the proofs, the straightforward part is to notice that every point in the left set lies in the right. But the less straightforward part lies in getting beyond the hunch that the superset could in fact have the same cardinality as $(0,1)$. (Of course, for this question one probably wouldn't think too much about the definition of cardinals like I did above.)

Indeed, to show that $[0,1)$ can be injected into $(0,1)$ one just needs to consider the injection given by

\begin{eqnarray*} f(0) & = & \frac{1}{2}\\ f(x) & = & x/2 ~~\text{if $x=2^{-k}$ for some $k\in\mathbb N$}\\ f(x) & = & x ~~\text{otherwise.} \end{eqnarray*}

Ok, time to get back to work.

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youngtableaux
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