# cardinality proof: prove that for any set $A, \ 2^{|A|} \neq |\mathbb{N}|$ [duplicate]

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 $$2^{|A|} = \aleph_{0}$$

• Do you know the definition of power set? Commented Feb 24, 2019 at 17:02
• @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). Commented Feb 24, 2019 at 17:07
• 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. Commented Feb 24, 2019 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 Commented Feb 24, 2019 at 17:15

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

• Therefore $|\mathcal{P}(A)|>|A|\ge\aleph_0$.
– J.G.
Commented Feb 24, 2019 at 17:07
• @J.G. Those inequalities you’ve mentioned seem off. Commented Feb 24, 2019 at 17:17
• They follow once you've precluded finite $|A|$, completing a proof by contradiction (relative to your old edit).
– J.G.
Commented Feb 24, 2019 at 17:33
• You’re correct on that end. The other part of the question is realizing $\aleph_{0}$ is the smallest infinite cardinal. Commented Feb 24, 2019 at 17:35
• Yeah, at some point you have to use that or prove it.
– J.G.
Commented Feb 24, 2019 at 17:52

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

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.

• What is $2^{|A|}-|\Bbb N|$? Commented Feb 24, 2019 at 19:29
• 2^|A|!=|N|; 2^|A|−|N|!=0, so used this as hypothesis. Commented Feb 26, 2019 at 6:18
• Cardinal subtraction is not well defined. Commented Feb 26, 2019 at 7:55
• please check the second proof Commented Feb 27, 2019 at 20:38