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Prove: for any set $A, \ 2^{|A|} \neq \aleph _{0}$

as $\aleph_{0} = |\mathbb{N}|$.

my attempt:

Suppose by contradiction that there exist a set $A$ such that $2^{|A|} = \aleph_{0}$, which Implies that $|P(A)| = \aleph_{0}$, using Cantor theorem.

meaning, intuitively that $ \aleph_{0} > |A| = \log_{2}(\aleph_{0})$, thus $A$ is finite.

at this point I'm trying to understand why the fact that $A$ is finite means that $P(A)$ also must be finite, which is the core of the argument.

there's no need to prove that $2^{|A|} = \mathfrak{c}$, but only to prove that there exist no set $A$ such that $2^{|A|} = \aleph_{0}$

Noa Even
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  • Do you know the definition of power set? – Cheerful Parsnip Feb 24 '19 at 17:02
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    @DietrichBurde I'd argue this is not a duplicate, since that result doesn't do the entirety of this question, and the result that $2^{\aleph_0} = c$ is stronger than required (we only need that it's uncountable, not that it's a particular other cardinality). – B. Mehta Feb 24 '19 at 17:07
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    The concept of $\log_2$ for infinite cardinals, in any standard sense, doesn't make any sense; the power-set operation isn't (necessarily) one-to-one on cardinalities. – Steven Stadnicki Feb 24 '19 at 17:09
  • @StevenStadnicki this is why I highlighted the word "intuitively" - which suggests that isn't a formal argument but only an intuition, as a part of the attempt to prove why there isn't exist a set which maintains this properties – Noa Even Feb 24 '19 at 17:15

2 Answers2

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If $A$ is finite, with $|A| = n$, then $|P(A)| = 2^{|A|} = 2^{n} < \aleph_{0}$ which is a contradiction to the definition of $\aleph_{0}$.

Noa Even
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Thomas Davis
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$|2^{|A|}-|\mathbb{N}|| \geq 2^{|A|}-|\mathbb{N}|$,

Now, $|A|<2^{|A|}$,

$2^{|A|}-|\mathbb{N}|>|A|-|\mathbb{N}|$,

If $2^{|A|} =|\mathbb{N}|$, then $|\mathbb{N}|>|A|$, meaning A is finite, so its power set will also be finite, contradicting the assumption. As, enter image description here

As long as $S$ is finite, series is convergent. Therefore, proved

Another proof,

Using 2^|A|=|N| as hypothesis, By Cantor–Bernstein–Schroeder theorem, 2^|A|=|N|

iff ,2^|A|>=|N| and 2^|A|<=|N|,

by Cantor's theorem, |A|<2^|A|

(|A|>=|N| or |A|<=|N|) and |A|<|N|,

(|A|>=|N| and |A|<|N|) or (|A|<=|N|and |A|<|N|),

the first curly bracket is a contradiction and second bracket refers A as finite set which is again an absurdity, so by logical disjuction, hypothesis of 2^|A|=|N| is false.

user121471
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