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This was a problem in the Scottish Book posed by Ulam. It was solved by

F. J. Freniche, The Number of Nonisomorphic Boolean subalgebras of a Power Set, Proc. Amer. Math. Soc., 91 (1984) 199-201.

the number of non isomorphic sub algebras is $2^{2^k}$ for any cardinal $k$

This was a problem in the Scottish Book posed by Ulam. It was solved by

F. J. Freniche, The Number of Nonisomorphic Boolean subalgebras of a Power Set, Proc. Amer. Math. Soc., 91 (1984) 199-201.

the number of non isomorphic sub algebras is $2^{2^k}$ for any infinite cardinal $k$

This was a problem in the Scottish Book posed by Ulam. It was solved by

F. J. Freniche, The Number of Nonisomorphic Boolean subalgefbras of a Power Set, Proc. Amer. Math. Soc., 91 (1984) 199-201.

the number of non isomorphic sub algebras is $2^{2^k}$ for any cardinal $k$

This was a problem in the Scottish Book posed by Ulam. It was solved by

F. J. Freniche, The Number of Nonisomorphic Boolean subalgebras of a Power Set, Proc. Amer. Math. Soc., 91 (1984) 199-201.

the number of non isomorphic sub algebras is $2^{2^k}$ for any cardinal $k$

This was a problem in the Scottish Book posed by Ulam. It was solved by

F. J. Freniche, The Number of Nonisomorphic Boolean subalgefbras of a Power Set, Proc. Amer. Math. Soc., 91 (1984) 199-201.

the number of non isomorphic sub algebras is $2^(2^k)$ for any cardinal $k$

This was a problem in the Scottish Book posed by Ulam. It was solved by

F. J. Freniche, The Number of Nonisomorphic Boolean subalgefbras of a Power Set, Proc. Amer. Math. Soc., 91 (1984) 199-201.

the number of non isomorphic sub algebras is $2^{2^k}$ for any cardinal $k$