Here's the question I am solving:
Let,
$P = [0, 1),QR = [0, 1) × [0, 1) = \{(x, y) ∈ \Bbb R^2: 0 ≤ x, y < 1\}$
Show that,
$card(P) = card(QR)$
This is what I have so far:
To show the sets P and QR are equal in cardinality, I aim to show that there is an injective function from $f: P→QR$ and a surjective function $g: QR→P$. This would mean there is a bijection and therefore $card(P) = card(QR)$
An injective function $y = (1-x)$ satisfies the given conditions and since there is an injective function $f: P→QR$ , it implies that $card(P)\le card(QR)$
I'm stuck trying to finding an example of a surjective function $g: QR→P$ that can show $card(P)\ge card(QR)$. Is the inverse function $x=(1-y)$ sufficient to show surjectivity?
I'm not certain of my answer, if it is correct and I'm just overthinking a trivial answer or if I'm trying to solve it incorrectly altogether.
In general, suppose we have sets $A, B$, with a surjective function $g : A \to B$ and an injective function $f : A \to B$.
The axiom of choice states (in one of several equivalent forms) that whenever $g : A \to B$ is a surjective function, there is a function $h : B \to A$ such that $g \circ h = 1_B$ - that is, such that for all $b \in B$, $g(h(b)) = b$.
This means that $h : B \to A$ is an injective function.
Since we have injective functions $f : A \to B$ and $h : B \to A$, we apply the Schroeder-Bernstein Theorem to conclude $|A| = |B|$.
However, this is not the tactic OP is using. OP is constructing an injective function $A \to B$ and a surjective function $B \to A$ [where $A = P$ and $B = QR$ in OP's case]. This is not enough to show that $|A| = |B|$.
Consider, for example, $A$ a 1-element set and $B$ a 2-element set. It's easy to see that both of the two functions $A \to B$ are injective, and also that the single function $B \to A$ is surjective. But $|A| \neq |B|$.