I'm aware that if there exists an injective map $f: A \to 2^A$ then for each element $a\in A$ $\exists$ $f(a)\subseteq A$.
But does this also mean $f(a)\subseteq 2^A$?
I ask this because when writing out a simple numerical example of such a set, say $A=\{{ 7,8\}}$ then $2^A=\{{\varnothing,\{{ 7\}},\{{ 8\}},\{{7,8\}}\}}$, however since each $f(a)$ is an element of $2^A$ then we must have $f(a)\subseteq 2^A$ since any set $X$ is always a subset of itself: $X\subseteq X$ right?
Thank you.
Best Regards.
To a question in title
No, $2^A \not\subseteq A$ because a power set of any $A$ has a cardinality strictly greater than $A$ itself.
To a question in text
No, $f(a)\subseteq 2^A$ does not hold in general, as $f(a)$ is by definition an element of $2^A$, and an element of a set is not (in general) a subset of the same set.
Althoug in some special cases it is, for example
$$\varnothing \in \{\varnothing, pink \} \text{ and } \varnothing \subset \{\varnothing, pink \}$$ however $$pink \in \{\varnothing, pink \} \text{ and } pink \not\subset \{\varnothing, pink \}$$