Prove that: $\aleph_0 \cdot \frak{c} = \frak{c} \cdot \frak{c}$

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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Is $C$ the cardinality of the continuum, $\frak{c} = 2^{\aleph_0}$? (\frak{c} gives $\frak{c}$) –  Arturo Magidin Aug 30 '11 at 16:17
is $C$ the cardinality of the reals? You can get that in TeX by typing \mathfrak{c}, $\mathfrak{c}$. –  Tyler Aug 30 '11 at 16:18
Yes thanks, fixed it –  Cu7l4ss Aug 30 '11 at 16:21
I assume that the answer that you found was mine. The reason you did not fully understand it is because the discussion was about infinite summation which is only equivalent to multiplication of two infinite cardinals in under some additional axioms. In my answer I'd shown that. The multiplication is just as easily defined with or without the additional axioms, and that is the essence of the first part of my answer there. –  Asaf Karagila Aug 30 '11 at 16:43
@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. –  Cu7l4ss Aug 30 '11 at 20:05

1 Answer

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

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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meaning I can find a bijection $f : \frak{c} \to \aleph_0 \times \frak{c}$ –  Cu7l4ss Aug 30 '11 at 16:23
Once you put this inequality together with the one you've established and Cantor-Bernstein, yes. –  Arturo Magidin Aug 30 '11 at 16:23
Damn, it has been right infront of me. Thank you very much! –  Cu7l4ss Aug 30 '11 at 16:25