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I am currently working on a proof regarding the cardinality of a set and its power set and would greatly appreciate some feedback and guidance.

Let $A$ be a set and let $P(A)$ denote its power set. I have defined a function $f: A \rightarrow P(A)$ such that $f(x) = \{x\}$ for every $x \in A$. It's clear that this function is injective. However, I noticed that this function is not surjective. For instance, for any two distinct elements $x, y \in A$, the set $\{x, y\}$ is in $P(A)$, but there doesn't exist an element $z \in A$ such that $f(z) = \{x, y\}$. So the function $f$ is not onto $P(A)$.

From this, I infer that the cardinality of $A$ is less than the cardinality of $P(A)$. My reasoning is that if a function from $A$ to $P(A)$ is not onto, then $P(A)$ must contain more elements than $A$.

Here are my questions:

Is my argument valid thus far?
How should I formally conclude my argument? Is it sufficient to say that since no function from $A$ to $P(A)$ can be onto, the cardinality of $P(A)$ must be greater than the cardinality of $A$?

Thank you for your time and help in advance.

Adam Rubinson
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WinnieXi
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  • Your argument of "less" suggests existence of two distinct elements in $A$ – Ivan Kaznacheyeu Jun 16 '23 at 12:59
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    A better justification for why $f$ is not surjective is that the empty set is in $P(A),$ but the empty is not in the image of $f,$ since every member of the image of $f$ has the form ${x},$ which is a set that contains something, rather than the empty set, which is a set that contains nothing /does not contain anything. – Adam Rubinson Jun 16 '23 at 13:01
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    Just saying that there is one function $A\to P(A)$ which is not surjective does not mean that every function $A\to P(A)$ is not surjective. You basically need to prove the latter claim. Indeed, if every function $A\to P(A)$ is not surjective, then there is no injection $P(A)\to A$ (I suppose this is using the axiom of choice). – Dave Jun 16 '23 at 13:08

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Your argument is not valid.

You cannot conclude $|A|<|P(A)|$ by constructing an injective and not surjective map from $A$ to P(A). For example, $f(n)=2n$ is a map from $\mathbb{Z}$ to $\mathbb{Z}$. It is injective and not surjective, but is $|\mathbb{Z}|<|\mathbb{Z}|$? Clearly not. It is only true when comparing the cardinality of finite sets. For infinite sets, the most useful method is to construct injective maps and use Bernstein theorem

You have proved that $|A|\leq |P(A)|$ by constructing an injective map. To prove $|A|<|P(A)|$, you can suppose there exists a bijective map and try to get a contradiction. See Cardinality of a set A is strictly less than the cardinality of the power set of A

PPP
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  • I see. In fact this was an idea that came up when reading the proof of the cardinal it’s theorem you mentioned. Thank you! – WinnieXi Jun 17 '23 at 13:04