Foundations of Advanced Mathematics

Let \(k : [0, ∞) \to [0, ∞)\) be defined by \(k(x) = x2\) for all \(x\in [0, ∞)\text{.}\) This function is surjective and injective, and hence bijective. First, we show that \(k\) is injective. Let \(x, y\in [0, ∞)\text{.}\) Suppose that \(k(x) = k(y)\text{.}\) Then \(x2 = y2\text{.}\) It follows that \(√x2 = √y2\text{,}\) and because \(x ≥ 0\) and \(y ≥ 0\text{,}\) we deduce that \(x = √x2 = √y2 = y\text{.}\) Hence \(k\) is injective. Second, we show that \(k\) is surjective. Let \(b\in [0, ∞)\text{.}\) Then \(√b\in [0, ∞)\text{,}\) and so \(k(√b) = (√b)2 = b\text{.}\) Hence k is surjective.

Let \(g : [0, ∞) \to R\) be defined by \(g(x) = x2\) for all \(x\in [0, ∞)\text{.}\) This function is injective but not surjective. The proof of the injectivity of \(g\) is the same as the proof of the injectivity of the function \(k\) in Part (1) of this example. The reason that \(g\) is not surjective is that \(g(a) \neq -2\) for any \(a\in [0, ∞)\text{,}\) though \(-2\) is in the codomain of \(g\text{.}\)

Let \(h : R \to [0, ∞)\) be defined by \(h(x) = x2\) for all \(x\in R\text{.}\) This function is surjective but not injective. The proof of the surjectivity of \(h\) is the same as the proof of the surjectivity of the function \(k\) in Part (1) of this example. The reason \(h\) is not injective is because \(h(-3) = 9 = h(3)\) even though \(-3 \neq 3\text{.}\) (Observe that instead of \(±3\) we could have used \(±a\) for any positive number \(a\text{,}\) but a single instance where the definition of injectivity fails is sufficient.)

Let \(f : R \to R\) be defined by \(f (x) = x2\) for all \(x\in R\text{.}\) This function is neither injective nor surjective, which is seen using the same arguments as the corresponding arguments for \(g\) and \(h\) in Parts (2) and (3) of this example.