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I've been fiddling with this enough. Found an answer here but didn't quite understand it.

How do I prove that: $\aleph_0 \cdot \frak{c} \leq \frak{c} \cdot \frak{c}$ $\frak{c} \cdot \frak{c} \leq \aleph_0 \cdot \frak{c}$

The first inequality I've proven already, the second one I just cannot get it(obvious is it may be).

The only thing I'm allowed to use is the definition of cardinality, the cantor-bernstein-schroder theorem and the fact that: $ k \leq p \wedge m \leq n \to k\cdot m \leq p \cdot n $ (which is helpful through the definition of cardinality)

Any insights are welcomed!

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    @Asaf Karagila Yes I've and I queried other answers too from that same question. The original problem is that I got stuck on using the cantor-bernstein theorem and not seeing other ways to prove it.2011-08-30

2 Answers 2

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$\mathfrak{c}\times\mathfrak{c} = 2^{\aleph_0}\times 2^{\aleph_0} = 2^{\aleph_0+\aleph_0} = 2^{\aleph_0} = 1\times 2^{\aleph_0} \leq \aleph_0\times 2^{\aleph_0}$.

For the equality $2^{\aleph_0}\times 2^{\aleph_0} = 2^{\aleph_0+\aleph_0}$, note that a pair of functions from $\aleph_0$ to $\{0,1\}$ is equivalent to a function from the disjoint union of two copies of $\aleph_0$ to $\{0,1\}$.

Since $\aleph_0+\aleph_0$ is the cardinality of the disjoint union of two copies of $\aleph_0=\omega$, and $2^{\kappa}$ is the cardinality of the set of all functions from $\kappa$ to $\{0,1\}$, the above argument seems to meet your requirements.

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    Damn, i$t$ has been right infront of me. Thank you very much!2011-08-30
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$\mathfrak{c}\times\mathfrak{c} =\text {max}\{\mathfrak{c},\mathfrak{c}\} =\mathfrak{c}=\text{max}\{\mathfrak{c},\aleph_0\}=\mathfrak{c}\times\aleph_0 $