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There are continuum many countable subsets of the continuum (because $\mathfrak{c}^{\aleph_0}=2^{\aleph_0}$). Thus the answer is $\mathfrak{c}$. See this questionthis question.
There are continuum many countable subsets of the continuum (because $\mathfrak{c}^{\aleph_0}=2^{\aleph_0}$). Thus the answer is $\mathfrak{c}$. See this question.
There are continuum many countable subsets of the continuum (because $\mathfrak{c}^{\aleph_0}=2^{\aleph_0}$). Thus the answer is $\mathfrak{c}$. See this question.
There are continuum many countable subsets of the continuum (because $\mathfrak{c}^{\aleph_0}=2^{\aleph_0}$). Thus the answer is $\mathfrak{c}$. See this question.
