Possible trivial counterexample to exam question So doing an exam question for set theory this was the question:
Let X be a set. Prove there is no injection $f : P(X) → X.$
Would X being an empty set be a counter example because in this case the powerset is a set of a single element $\{\}$ hence it is mapped to a null element. 
However would this be the case that $f$ would not be a function, since there isn't $(a,b)\in f$ hence $f$ isn't defined for the whole of the domain hence $f$ is not a function.
So yes I understand that this isn't a proof for this problem but while thinking about the problem I was trying some examples and non-examples and couldn't work out why this failed until I actually put it into words.
Sorry for a convoluted question, I don't really have the correct notation to explain myself.
 A: No, $X=\varnothing$ is not a counterexample. You seem to have realized (though your wording is a bit strange) that there is no function $\mathcal P(\varnothing)\to\varnothing$ -- so in particular there is no injective function, so we have seen that the claim is true for $X=\varnothing$.
(A "counterexample" means an example that makes the claim false).

Any course that has exam questions of this kind will probably have proved the celebrated

Cantor's Theorem. Let $X$ be a set. No function $X\to\mathcal P(X)$ can be surjective.

So one strategy would be to show that if there were an injection $\mathcal P(X)\to X$ you could construct a surjection $X\to \mathcal P(X)$.
Another way would be to combine the assumed injection $\mathcal P(X)\to X$ with the obvious injection $X\to\mathcal P(X)$, so the Cantor-Schröder-Bernstein theorem would conclude that there is a bijection $X\to\mathcal P(X)$, which again contradicts Cantor's theorem.
A: One can "dualise" the proof of Cantor's theorem to the present situation:
If $f:\>{\cal P}(X) \to X$ is injective with image set $P\subset X$ then we have a well defined  inverse map $f^{-1}:\>P\to{\cal P}(X)$. Consider the "special" set $A:=\{x\in P\,|\, x\notin f^{-1}(x)\}$, and put $a:=f(A)\in P$. Then it is easy to check that one has
$$a\in A\quad\Leftrightarrow\quad a\notin f^{-1}(a)=A\ ,$$
which is absurd.
