Prove that $2^\aleph+\aleph$ equals $2^\aleph$.

Question

How do I show this elegantly? I can't seem to find the right sets for it... Maybe there are some substantial laws I could use? I would appreciate your help...

What I said is: $2^\mathfrak c$ is for sure greater than $\mathfrak c$. Furthermore, $$2^\mathfrak c\cdot 2^\mathfrak c=2^\mathfrak c+\mathfrak c = 2^\mathfrak c.$$ On the other hand, $$2^\mathfrak c\cdot 2^\mathfrak c= 2^\mathfrak c+\underbrace{(2^\mathfrak c+\ldots+2^\mathfrak c)}_{2^\mathfrak c -1\text{ times}}$$ which is greater than $2^\mathfrak c+(\mathfrak c+\ldots+\mathfrak c)$ ($2^\mathfrak c - 1$ times) which is greater then $2^\mathfrak c+\mathfrak c$, which means $2^\mathfrak c$ is larger then $2^\mathfrak c+\mathfrak c$.

Together with $2^\mathfrak c+\mathfrak c$ being larger then $2^\mathfrak c$, we get, by Cantor-Schroeder-Bernstein theorem that $2^\mathfrak c+\mathfrak c=2^\mathfrak c$.

Any corrections?

Answer 1

$$ 2^\aleph = 2^{\aleph+1} = 2 \cdot 2^\aleph = 2^ \aleph + 2^\aleph \ge 2^ \aleph + \aleph \ge 2^ \aleph $$

and this proves from Cantor–Schroeder–Bernstein theorem.

Answer 2

Your edit makes the question trivial for infinite $\aleph$. It remains false for finite $\aleph$.

Answer 3

On the one hand, $2^\aleph\leq\aleph+2^\aleph$. On the other hand, you can show quite easily that because $\aleph+\aleph=\aleph$ we have that $2^\aleph\cdot 2^\aleph=2^\aleph$, and therefore $2^\aleph+2^\aleph=2^\aleph$ as well.

Now using Cantor-Bernstein you have $$\aleph+2^\aleph\leq2^\aleph+2^\aleph=2^\aleph$$

Note that proving this with specific sets is not elegant. It's never elegant. Cardinal arithmetic is the elegant part here, and proving the first volley of theorems of cardinal arithmetic is rarely "elegant", and mostly exercise in composing bijections and injections. But then you get to use it for particular cases like here, and end up with quick proofs for something that is otherwise tedious and annoying.

As for what you have written, while not incorrect, it requires the use of the axiom of choice to prove that cardinal multiplication is just like repeated addition infinitely many times; it is also considered the far less elegant solution. Your mind is in the right place, but the execution is lacking. While for finite cardinals we can, and sometimes should, replace the product by sums it is rarely the case for infinite cardinals.