# Constructing a Bijection

• May 18th 2008, 04:56 PM
ThePerfectHacker
Constructing a Bijection
A problem in my book states to prove $n\cdot \aleph_{\alpha} = \aleph_{\alpha}$ where $n$ is a positive integer.
This problem requires to construct a one-to-one mapping from $\omega_{\alpha} = n\times \omega_{\alpha}$.

I think I found a way to do this using transfinite recursion however it is not explicit. Is that good enough? Because recursion is not actually explicit. Is there an actual way of constructing an explicit correspondence from $\omega_{\alpha}$ to $n\times \omega_{\alpha}$? My intuition begins to break down when dealing with very large sets.

Here is how I was doing it. This is the following transfinite recusion theorem I will use: let $\Omega$ be an ordinal and $A$ a set, given $g: \bigcup_{\alpha < \Omega}A^{\alpha}\mapsto A$ there exists (a unique) function $f: \Omega \mapsto A$ so that $f(\alpha) = g(f|\alpha)$ for all $\alpha < \Omega$.

We begin by well-ordering $n\times \omega_{\alpha}$ lexiographically. Let $S = \bigcup_{\beta < \omega_{\alpha}}(n\times \omega_{\alpha})^{\beta}$. Let $a\not \in n\times \omega_{\alpha}$. Now define a function $g: S\mapsto n\times \omega_{\alpha}$ as $g(x) =\left\{\begin{array}{c} \mbox{least element of } n\times \omega_{\alpha} - \mbox{ran}(x) \mbox{ if }\not = \emptyset \\ a \mbox{ if }=\emptyset \end{array} \right.$. By transfinite recursion there exists a function $f:\omega_{\alpha} \mapsto n\times \omega_{\alpha}$ so that $f(\beta) = g(f|\beta) = \left\{ \begin{array}{c} \mbox{least element of } n\times \omega_{\alpha} - \mbox{ran}(f|\beta) \mbox{ if }\not = \emptyset \\ a\mbox{ if }=\emptyset \end{array} \right.$.

To complete the proof that $f$ provides a bijection we need to do show a few more details. One of them is that $f$ must attain a value of $a$ at some point. Because otherwise we would have an injection from $\omega_{\alpha}$, and arrive at a contradiction from there.

I just do not like this approach because it is not really an explicit mapping.