# Proof

(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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### References

#### Bibliography

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