Posted by haifeng on 2011-07-03 09:10:13 last update 2011-08-24 17:52:58 | Edit | Answers (1)
1
Posted by haifeng on 2011-08-24 17:51:25
\[2^{\aleph_0}\leqslant\Pi_{i=1}^{+\infty}i\leqslant{\aleph_0}^{\aleph_0},\]
注意 $c=2^{\aleph_0}={\aleph_0}^{\aleph_0}$.