Cardinal Numbers
Solved Problems
Example 1.
Find \({\aleph_0}^{{\aleph_0}}.\)
Solution.
By Cantor's theorem, we have \({\aleph_0} \lt {2^{{\aleph_0}}}.\) Then
\[{\aleph_0}^{{\aleph_0}} \le {\left( {{2^{{\aleph_0}}}} \right)^{{\aleph_0}}} = {2^{{\aleph_0} \cdot {\aleph_0}}} = {2^{{\aleph_0}}}.\]
From the other side, \({2^{{\aleph_0}}} \le {\aleph_0}^{{\aleph_0}}.\)
Therefore, using the Cantor-Schröder-Bernstein Theorem and assuming the continuum hypothesis, we get
\[{\aleph_0}^{{\aleph_0}} = {2^{{\aleph_0}}} = {\aleph_1}.\]
Example 2.
Prove that \({{\aleph_0}^{{\aleph_i}}} = {\aleph_{i + 1}}.\)
Solution.
Since \({\aleph_0} \lt {2^{{\aleph_0}}}\) by Cantor's theorem, we have
\[{\aleph_0}^{{\aleph_i}} \le {\left( {{2^{{\aleph_0}}}} \right)^{{\aleph_i}}} = {2^{{\aleph_0} \cdot {\aleph_i}}} = {2^{\max \left( {{\aleph_0},{\aleph_i}} \right)}} = {2^{{\aleph_i}}} = {\aleph_{i + 1}}.\]
On the other hand, \(\aleph_0 \gt 2,\) so
\[{\aleph_0}^{{\aleph_i}} \ge {2^{{\aleph_i}}} = {\aleph_{i + 1}}.\]
By the Cantor-Schröder-Bernstein Theorem, we obtain
\[{\aleph_0}^{{\aleph_i}} = {\aleph_{i + 1}}.\]
Example 3.
Prove that \({n^{{\aleph_i}}} = {\aleph_{i + 1}},\) where \(n \gt 1\) is a finite natural number.
Solution.
Since \(n \lt 2^n,\) we can write
\[{n^{{\aleph_i}}} \le {\left( {{2^n}} \right)^{{\aleph_i}}} = {2^{n \cdot {\aleph_i}}} = {2^{\max \left( {n,{\aleph_i}} \right)}} = {2^{{\aleph_i}}} = {\aleph_{i + 1}}.\]
At the same time, since \(n \ge 2,\) we have
\[{n^{{\aleph_i}}} \ge {2^{{\aleph_i}}} = {\aleph_{i + 1}}.\]
According to the Cantor-Schröder-Bernstein Theorem, this means
\[{n^{{\aleph_i}}} = {\aleph_{i + 1}}.\]
For example,
\[{5^{{\aleph_i}}} = {\aleph_{i + 1}},\;\;{7^{{\aleph_i}}} = {\aleph_{i + 1}}.\]
Example 4.
Find the cardinality of the set of all functions from \(\mathbb{R}\) to \(\mathbb{R}\).
Solution.
The set of all functions \(f: \mathbb{R} \to \mathbb{R}\) is given by \(\mathbb{R}^\mathbb{R},\) so the cardinality of this set is
\[\left| {{\mathbb{R}^\mathbb{R}}} \right| = {\mathfrak{c}^\mathfrak{c}}.\]
Since \(\mathfrak{c} = {2^{{\aleph_0}}},\) we have
\[\left| {{\mathbb{R}^\mathbb{R}}} \right| = {\mathfrak{c}^\mathfrak{c}} = {\left( {{2^{{\aleph_0}}}} \right)^\mathfrak{c}} = {2^{{\aleph_0}\mathfrak{c}}}.\]
We know that
\[{\aleph_0}\mathfrak{c} = \max \left( {{\aleph_0},\mathfrak{c}} \right) = \mathfrak{c}.\]
Hence,
\[\left| {{\mathbb{R}^\mathbb{R}}} \right| = {\mathfrak{c}^\mathfrak{c}} = {2^{{\aleph_0}\mathfrak{c}}} = {2^c}.\]
Assuming the extended continuum hypothesis, we can represent the answer as follows:
\[\left| {{\mathbb{R}^\mathbb{R}}} \right| = {2^\mathfrak{c}} = {2^{{\aleph_1}}} = {\aleph_2}.\]
Example 5.
Simplify the expression
\[\left( {3\aleph_0^2 + {\mathfrak{c}^{{\aleph_0}}}} \right)\left( {{3^{2{\aleph_0}}} + {\mathfrak{c}^2}} \right).\]
Solution.
We calculate each term separately.
\[3\aleph_0^2 = \max \left( {3,{\aleph_0},{\aleph_0}} \right) = {\aleph_0};\]
Assuming the continuum hypothesis, we have
\[{\mathfrak{c}^{{\aleph_0}}} = {\left( {{2^{{\aleph_0}}}} \right)^{{\aleph_0}}} = {2^{{\aleph_0} \cdot {\aleph_0}}} = {2^{\max \left( {{\aleph_0},{\aleph_0}} \right)}} = {2^{{\aleph_0}}} = \mathfrak{c} = \aleph_1;\]
\[{3^{2{\aleph_0}}} = {3^{\max \left( {2,{\aleph_0}} \right)}} = {3^{{\aleph_0}}} = {\aleph_1}\;\left({\text{see Example }3}\right);\]
\[{\mathfrak{c}^2} = {\left( {{2^{{\aleph_0}}}} \right)^2} = {2^{2{\aleph_0}}} = {2^{\max \left( {2,{\aleph_0}} \right)}} = {2^{{\aleph_0}}} = \mathfrak{c} = {\aleph_1}.\]
Hence,
\[\left( {3\aleph_0^2 + {c^{{\aleph_0}}}} \right)\left( {{3^{2{\aleph_0}}} + {c^2}} \right) = \left( {{\aleph_0} + {\aleph_1}} \right)\left( {{\aleph_1} + {\aleph_1}} \right) = \max \left( {{\aleph_0},{\aleph_1}} \right) \cdot \max \left( {{\aleph_1},{\aleph_1}} \right) = {\aleph_1} \cdot {\aleph_1} = \max \left( {{\aleph_1},{\aleph_1}} \right) = {\aleph_1}.\]
Example 6.
Simplify the expression
\[{3^{{\aleph_2}}} \cdot \aleph_4^3 \cdot {\aleph_2}^{{\aleph_4}} \cdot {\aleph_4}^{{\aleph_2}.}\]
Solution.
We compute each factor separately.
\[{3^{{\aleph_2}}} = {\aleph_3}\;\left({\text{see Example }3}\right);\]
\[\aleph_4^3 = \max \left( {{\aleph_4},{\aleph_4},{\aleph_4}} \right) = {\aleph_4};\]
\[{\aleph_2}^{{\aleph_4}} = {\left( {{2^{{\aleph_1}}}} \right)^{{\aleph_4}}} = {2^{{\aleph_1} \cdot {\aleph_4}}} = {2^{\max \left( {{\aleph_1},{\aleph_4}} \right)}} = {2^{{\aleph_4}}} = {\aleph_5};\]
\[{\aleph_4}^{{\aleph_2}} = {\left( {{2^{{\aleph_3}}}} \right)^{{\aleph_2}}} = {2^{{\aleph_3} \cdot {\aleph_2}}} = {2^{\max \left( {{\aleph_3},{\aleph_2}} \right)}} = {2^{{\aleph_3}}} = {\aleph_4}.\]
Therefore,
\[{3^{{\aleph_2}}} \cdot \aleph_4^3 \cdot {\aleph_2}^{{\aleph_4}} \cdot {\aleph_4}^{{\aleph_2}} = {\aleph_3} \cdot {\aleph_4} \cdot {\aleph_5} \cdot {\aleph_4} = \max \left( {{\aleph_3},{\aleph_4},{\aleph_5},{\aleph_4}} \right) = {\aleph_5}.\]
