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Let $A$ a infinite set, prove $card(A)+card(A)=card(A)$

My attempt:
Let $A$ a infinite set. Then we have a subset $B$ of $A$ such that B is infinte countable, this implies: $card(B)=card(\mathbb{N})$

Then: $card(\mathbb{N})=card(\mathbb{N}\cup\mathbb{N})=card(\mathbb{N})+card(\mathbb{N})$.

This implies:

$card(B)=card(B)+card(B)$

But i need this for $card(A)$. Here i'm stuck, can someone help me?

Andrés E. Caicedo
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    Instead of relying on the fact that $|\mathbb{N}| + |\mathbb{N}| = |\mathbb{N}|$, perhaps you could repeat the proof of the fact, using $|A|$ instead? – Nicholas Viggiano Nov 04 '19 at 20:07
  • @NicholasViggiano Do you have a simple proof of this? I imagine you well-order $A$, then... ? – diracdeltafunk Nov 04 '19 at 20:12
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    @diracdeltafunk Do the usual trick to replace the two copies of $A$ with disjoint sets, ie set the first to $A \times {0}$ and the second to $A \times {1}$, and take the cardinality of their union. Then, yes, well-order $A$ and alternately send the elements to the two. – Nicholas Viggiano Nov 04 '19 at 20:23
  • @NicholasViggiano I was a little tripped up about how to assign a "parity" to each limit ordinal, but I think it's fine to just say they're all "even". Thanks! – diracdeltafunk Nov 04 '19 at 20:32

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Instead of finding just one $B \subseteq A$ such that $\operatorname{Card}(B) = \operatorname{Card}(\mathbb{N})$, you want to partition $A$ into subsets of cardinality $\aleph_0 = \operatorname{Card}(\mathbb{N})$. Hint: use Zorn's Lemma.

Once you've done this, you'll have $A \cong \coprod_{i \in I} B_i$ for some collection of countably infinite sets $\{B_i : i \in I\}$ (whence $B_i \cong B_i \amalg B_i$ as you noted). Then

$$A \amalg A \cong \left(\coprod_{i \in I} B_i\right) \amalg \left(\coprod_{i \in I} B_i\right) \cong \coprod_{i \in I} (B_i \amalg B_i) \cong \coprod_{i \in I} B_i \cong A.$$

diracdeltafunk
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