# Cardinality of all cardinal numbers less than a given cardinal

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 $$X_{\alpha?}$$

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