0
$\begingroup$

For a given cardinal number $\aleph_{\alpha}$ we define $$X_{\alpha}= \{\aleph_{\beta}; \aleph_{\beta}<\aleph_{\alpha}\}.$$ We can easily prove that

1) $card(X_{\alpha}) \leq \aleph_{\alpha}^{+}=2^{\aleph_{\alpha}},$ and

2) if $\alpha$ is a successor cardinal number, then $card(X_{\alpha}) \leq \aleph_{\alpha}.$

Now, can we show that $card(X_{\alpha}) \leq \aleph_{\alpha}$ in general? What we can say about the cardinality of $X_{\alpha?}$

$\endgroup$
  • 5
    $\begingroup$ $|X_\alpha|=|\alpha|$ surely? $\endgroup$ – Lord Shark the Unknown Feb 13 at 7:46
  • $\begingroup$ Is there any reason? $\endgroup$ – Ali Bayati Feb 13 at 7:52
  • 5
    $\begingroup$ There exists a natural bijection $\alpha \rightarrow X_{\alpha}$, which is $\beta \longmapsto \aleph_{\beta}$. $\endgroup$ – Mindlack Feb 13 at 8:11
  • $\begingroup$ at most X_alpha. $\endgroup$ – Jacob Wakem Feb 13 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.