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.
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