(related to Lemma: Invertible Functions on Real Intervals)

The proof involves three steps:

\((1)\) Showing that \(f\) is invertible.

By hypothesis, \([a,b]\) is a closed real interval and \(f:[a,b]\to\mathbb R\) is a strictly monotonically increasing real function. (The proof is analogue for strictly monotonically decreasing functions). Therefore, for an \(x\) with \(a < x < b\) we have \(A:=f(a) < f(x) < B:=f(b)\). Therefore, \(f\) is a function mapping the closed interval \([a,b]\) to the closed interval \([A,B]\). Because from \( x < x'\) it follows that \(f(x) < f(x')\), this function is injective. Because \(f\) is a continuous real function, it follows from the intermediate value theorem that \(f\) takes any value between \(A\) and \(B\). Thus, \(f\) is also surjective. Because \(f\) is both, injective and surjective, it is bijective. As a bijective function, an inverse function \(f^{-1}\) must exist, thus \(f\) is invertible.

\((2)\) Demonstrating that the inverse function \(f^{-1}\) is strictly monotonically increasing.

According to \((1)\), it follows from \( x < x'\) that \(f(x) < f(x')\). Analogously, we can conclude from \(f(x) < f(x') \) that \(f^{-1}\circ f(x)= x < f^{-1}\circ f(x')=x'\). (The proof is analogue for strictly monotonically decreasing functions).

\((3)\) Proving that the inverse function \(f^{-1}\) is continuous.

The proof will be by contradiction. Suppose, \(y\in[A,B]\) and \((y_n)_{n\in\mathbb N}\) is real sequence with \(y_n\in[A,B]\) and \(\lim_{n\to\infty} y_n=y\), but \(\lim_{n\to\infty} f^{-1}(y_n)\neq f^{-1}(y)\). We will show that this cannot be true, i.e. it must be \(\lim_{n\to\infty} f^{-1}(y_n)=f^{-1}(y)\), which is equivalent to \(f^{-1}\) being continuous.

\(\lim_{n\to\infty} f^{-1}(y_n)\neq f^{-1}(y)\) would imply that there exists an \(\epsilon > 0\) with \[|f^{-1}(y_n)-f^{-1}(y)| \ge \epsilon\] for infinitely many \(n\). Therefore there exists a subsequence \((y_{n_k})_{k\in\mathbb N}\) with \[|f^{-1}(y_{n_k})-f^{-1}(y)| \ge \epsilon\] for all \(k\in\mathbb N\). Because by definition \(a < f^{-1}(y_{n_k})_{k\in\mathbb N} < b\), the sequence \((f^{-1}(y_{n_k}))_{k\in\mathbb N}\) is bounded, thus it contains another subsequence, which is convergent due to the theorem of Bolzano-Weierstrass. Let's call this other subsequence \((f^{-1}(y_{n_j}))_{j\in\mathbb N}\) and let's assume \(\lim_{j\to\infty} f^{-1}(y_{n_j})=c\). Note that, since convergence preserves upper and lower bounds, it must follow from \[|f^{-1}(y_{n_j})-f^{-1}(y)| \ge \epsilon\] for all \(j\in\mathbb N\) that \[|c-f^{-1}(y)| \ge \epsilon\quad\quad( * ).\] Now, note that \(f(f^{-1}(y_{n_j}))=y_{n_j}\) and that from the continuity of \(f\) it follows \[y=\lim_{n\to\infty}y_n=\lim_{j\to\infty}f(f^{-1}(y_{n_j}))=f(\lim_{j\to\infty}f^{-1}(y_{n_j}))=f( c).\] This would mean that \(f^{-1}(y)=f^{-1}(f( c))=c\), which is a contradiction to \( ( * )\). Thus, our assumption must be wrong and \(f^{-1}\) must be continuous.

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  1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983