I'm struggling with this question (The function $f(x) = x^2 -3$):
Let $A = \{x \in R : x \geq 0\}$. Determine a maximum set $B$ such that $f : A \rightarrow B$ is a bijection. Let $g : B \rightarrow A$ be a function defined by $g(x) = \sqrt{x + k}$ for some fixed real number $k \geq 3$. Determine $f \circ g$ and $g \circ f$.
A maximal set $B$ has to have greater cardinality, and not be a subset of $A$ I thought. But $R$ doesn't have cardinality $\aleph_0$, and I'm not sure how to make a bigger set?
The question seems garbled, but I'll try to clear it up. We're given a function $f\colon x\mapsto x^2 - 3, \colon A\to \Bbb R$. defined on $A=\{x\mid x\ge 0\}$, the nonnegative reals. So $A = \Bbb R_{\ge 0}$.
On $\Bbb R_{\ge 0}$, $f$ sends $0$ to $-3$, it's strictly increasing, unbounded, continuous, and more. It's an injection, with range $B=\{x\mid x\ge -3\}$. This $B$ is the only set that $f$ on $A$ can be a bijection to. You can't shrink B, and you can't enlarge it to $B'\supset B = f(A)$ and still have $f$ be a bijection to $B'$. There aren't several such $B$, there's only that $B$.
This talk of the "maximum such $B$" seems misguided and misleading when talking about the range of the function. The maximum such $B$ as well as the minimum exists, but both are equal to the only such $B$.
Consider the domain of $f$, though. Certainly $f$ can be extended to all of the reals, therefore to any set of reals $A'$ containing $A$. But if $A\subsetneqq A' \subseteq \Bbb R$, then $A' \!\setminus\! A$ contains a negative number $x_0 < 0$. Therefore $0 < -x_0$, so $-x_0\in A\subseteq A'$. Now, $f$ is an even function: for all $x$ in its domain, $f(-x) = f(x)$. Thus $f(x_0) = f(-x_0)$. So $f$ is not an injection on $A'$.
When considering the domain of $f$, as we just did, it turns out to be very meaningful to ask, Is there a maximum subset $A$ to which $f$ can be extended and still be an injection, and if so, what is it? There is, and it's $\Bbb R_{\ge 0}$.