Suppose that f : X --> Y is given.
If X is finite, show that f(X) is also finite. Similarly, if X is countable,
show that f(X) is also countable.
Hello,
(talking about Y is an extraneous information)
The solution lies in the definition of $\displaystyle f(X)$
$\displaystyle f(X)=\{f(x) ~:~ x \in X\}$
For any $\displaystyle x \in X$, there is at least one element $\displaystyle y \in f(X)$ such that $\displaystyle y=f(x)$
It also means that for any $\displaystyle y \in f(X)$, there is at least one $\displaystyle x \in X$ such that $\displaystyle y=f(x)$
You can see that this sentence is exactly the definition of a surjection.
Hence $\displaystyle f ~:~ X \to f(X)$ is a surjection. By Cantor-Bernstein theorem (I think it's this one), we have $\displaystyle \text{Card}(X) \ge \text{Card}(f(X))$
If X is finite, then $\displaystyle \text{Card}(X) < \text{Card}(\mathbb{N})$. So we have $\displaystyle {\color{red}\text{Card}(f(X))} \le \text{Card}(X) < {\color{red}\text{Card}(\mathbb{N})}$. Thus f(X) is finite.
If X is countable, then $\displaystyle \text{Card}(X) \le \text{Card}(\mathbb{N})$, etc...
If you see any mistake, tell me, I'm learning this stuff too
Here are some remarks to clear up some questions raised by the above posts.
DEFINITION: $\displaystyle card(A) \leqslant card(B) \Leftrightarrow \,\exists \, g:A \mapsto B$ is an injection
Here is a most important theorem in many texts that deal with this material.
There is an injection $\displaystyle F:A \mapsto B$ if and only if there exists a surjection $\displaystyle
G:B \mapsto A$.
As noted above there is a surjection $\displaystyle f:X \mapsto f(X)$.
So by that theorem we must have an injection $\displaystyle g:f(X) \mapsto X$
Now apply the definition.
BTW: I do not see how the Cantor-Schroder-Bernstein theorem applies
Also recall that a function $\displaystyle f:A \mapsto B$ is simply a subset of $\displaystyle A \times B$.
Where $\displaystyle A$ is the set of all first terms of the pairs and $\displaystyle f(A)$ is the set of second terms in the pairs.
So there is a natural expectation that the $\displaystyle card(f(A)) \leqslant card(A)$.