Haudorff Formula Set Theory

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For every $\alpha$ and every $\beta$, $$\aleph_{\alpha+1}^{\aleph_{\beta}}=\aleph_{\alpha}^{\aleph_{\beta}} \cdot \aleph_{\alpha+1}$$

Proof: If $\beta \geq \alpha+1$, then $\aleph_{\alpha+1}^{\aleph_{\beta}}=2^{\aleph_{\beta}}, \aleph_{\alpha}^{\aleph_{\beta}}=2^{\aleph_{\beta}}$, and $\aleph_{\alpha+1} \leq \aleph_{\beta} \leq 2^{\aleph_{\beta}}$. Hence, the formula holds. Thus let us assume that $\beta \leq \alpha$. Since $\aleph_{\alpha}^{\aleph_{\beta}} \leq \aleph_{\alpha+1}^{\aleph_{\beta}}$ and $\aleph_{\alpha+1} \leq \aleph_{\alpha+1}^{\aleph_{\beta}}$, it suffices to show that $\aleph_{\alpha+1}^{\aleph_{\beta}} \leq \aleph_{\alpha}^{\aleph_{\beta}} \cdot \aleph_{\alpha+1}$.

Each function $f : \omega_{\beta} \rightarrow \omega_{\alpha+1}$ is bounded; i.e. there is $\gamma < \omega_{\alpha+1}$ such that $f(\xi) < \gamma$ for all $\xi < \omega_{\beta}$ (this is because $\omega_{\alpha+1}$ is regular and $\omega_{\beta} < \omega_{\alpha+1}$). Hence,

$$\omega_{\alpha+1}^{\omega_{\beta}}=\bigcup_{\gamma < \omega_{\alpha+1}}\gamma^{\omega_\beta}$$

Now every $\gamma < \omega_{\alpha+1}$ has cardinality $|\gamma| \leq \aleph_{\alpha}$, and we have $|\bigcup_{\gamma < \omega_{\alpha+1}}\gamma^{\omega_{\beta}}| \leq \sum_{\gamma < \omega_{\alpha+1}}|\gamma|^{\aleph_\beta}$. Thus,

$$\aleph_{\alpha+1}^{\aleph_{\beta}} \leq \sum_{\gamma < \omega_{\alpha+1}}|\gamma|^{\aleph_{\beta}} \leq \sum_{\gamma < \omega_{\alpha+1}}\aleph_{\alpha}^{\aleph_{\beta}}=\aleph_{\alpha}^{\aleph_{\beta}} \cdot \aleph_{\alpha+1}$$

Questions:

1) After the first sentence, how we conclude the formula holds?

2) Do we need regularity of $\omega_{\alpha+1}$ to conclude $f$ is bounded?

3) How to obtain the equality $$\omega_{\alpha+1}^{\omega_{\beta}}=\bigcup_{\gamma < \omega_{\alpha+1}}\gamma^{\omega_\beta}$$

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  1. Because $\aleph_{\alpha+1}^{\aleph_\beta}=2^{\aleph_\beta}$.

  2. Because $\kappa$ is a singular cardinal exactly when there is an unbounded function from some $\lambda<\kappa$ to $\kappa$.

  3. Because each function is bounded, each function is in fact from $\omega_\beta$ into some $\gamma<\omega_{\alpha+1}$ (e.g. $\gamma$ taken as the successor of the supremum of the range)