0

Let $ \aleph_{\alpha} $ be a cardinal and assume that $ \left\{ A_{\beta}:\beta<\aleph_{\alpha}\right\} $ is a set of sets, such that $ |A_{\beta}|=\aleph_{\alpha} $ for any $ \beta<\aleph_{\alpha} $.

Prove that exists set of sets $ \left\{ D_{\beta}:\beta<\aleph_{\alpha}\right\} $ such that :

  1. $ D_{\beta}\subseteq A_{\beta} $

  2. $ |D_{\beta}|=\aleph_{\alpha} $

  3. if $ \beta\neq\gamma $ then it follows that $ D_{\beta}\cap D_{\gamma}=\emptyset $.

So, actually what I need to do, is to choose different values $ \aleph_{\alpha} $ from each $ A_{\alpha} $

Its a bit complicated for me. If the task would be to choose just one different value from each set, then I would know how to solve it. But here I have to choose $ \aleph_{\alpha} $ values and make sure that they are different. So any ideas would be highly appreciated.

FreeZe
  • 3,735

2 Answers2

2

Hint: You need to pick a total of $\aleph_\alpha\cdot\aleph_\alpha=\aleph_\alpha$ elements to form all the sets $D_\beta$. Pick them one by one in a transfinite recursion of length $\omega_\alpha$.

More details are hidden below.

Let $f=(f_0,f_1):\omega_\alpha\to\omega_\alpha\times\omega_\alpha$ be a bijection. The idea is then we do a recursion of length $\omega_\alpha$ such that in the $\beta$th step, we pick the $f_1(\beta)$th element of $D_{f_0(\beta)}$. More precisely, recursively define a sequence $(x_\beta)_{\beta<\omega_\alpha}$ such that each $x_\beta$ is an element of $A_{f_0(\beta)}$ and is different from $x_\gamma$ for all $\gamma<\beta$. We can do this because $|A_{f_0(\beta)}|=\aleph_\alpha$ and there are only $|\beta|<\aleph_\alpha$ such $x_\gamma$. Now take $D_\beta=\{x_\gamma:f_0(\gamma)=\beta\}$.

Eric Wofsey
  • 330,363
0

Defining ordered pairs e.g. as thus, and functions as thus, and using bijections to impose certain helpful assumptions about the structure of the $A_\beta$ (you can figure out the details as an exercise), we can proceed as follows. Without loss of generality $A_\beta$ is a function whose range is a subset of its domain, which is $\omega_\alpha^2$. Define $D_\beta$ as $A_\beta$ restricted to the domain $\{\omega_\alpha\cdot\beta+\gamma|\gamma<\omega_\alpha\}$.

J.G.
  • 115,835