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There is a theorem on continuous function that goes as follow:

If f is a continuous strictly monotone function defined on an interval, then its inverse is also a continuous function.

I have quite an ugly proof on this theorem. My textbook proof doesn't look good either. So I am just wondering if someone can provide me with a more elegant proof. Thanks.

On top of that, I am just wondering why must the function be strictly monotone? Is it to ensure that it is one-to-one so that the inverse exists? Thanks.

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  • $\begingroup$ You are exactly right about the reason for strict monotonicity. $\endgroup$ – Cameron Buie Jul 29 '15 at 0:23
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The statement looks more "obviously true" in the view of topology. Suppose $f$ is defined on an open interval $(a,b)$. (If not, for example, $f$ as a continuous strictly increasing function is defined on $[a,b]$, then by putting $F(x) = f(x), x \in [a,b]$, $F(x) = x + f(a) - a, x \in (a-1,a)$ and $F(x) = x + f(b) - b, x \in (b, b+1)$, $f$ is extended to an open interval without losing monotonicity). Besides, such extension is not necessary but for convenient expression of the basis.

Note that $\mathscr{B} = \{(r_1,r_2): a < r_1 < r_2 < b \}$ is a basis of the topology on $(a,b)$. By continuity we have $f: (a,b) \rightarrow \big(f(a), f(b) \big)$ is one-to-one, and $f\big((r_1, r_2)\big) = (f(r_1), f(r_2))$ is open in the topological space $\big( f(a), f(b) \big)$, so $f$ is open mapping and thus $f^{-1}$ is continuous.

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    $\begingroup$ In fact, you can omit continuity of $f$ and still get the same conclusion. See part 2 of Alan Keith Austin, Two curiosities, Mathematical Gazette 69 #447 (March 1985), 42-44. $\endgroup$ – Dave L. Renfro Apr 30 '14 at 13:26
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If a proof with elementary arguments is desired, here's a simple one:

Suppose first that $f$ is strictly increasing on $I=(a,b)$, and $y\in f(I)$, with $f^{-1}(y)=x.$

$\forall \epsilon>0$, let $\epsilon'=\min\{\epsilon, \frac{b-x}{2}, \frac{x-a}{2}\}.$ Suppose $f(x+\epsilon')=y+\delta_1$ and $f(x-\epsilon')=y-\delta_2$. Since $f$ is strictly increasing, clearly $f^{-1}(t)<x+\epsilon'$ if $t<y+\delta_1$. Similarly, $f^{-1}(t)>x-\epsilon'$ if $t>y-\delta_2.$ (Obviously $t$ also has to be in $f(I)$.) Now, with $\delta=\min(\delta_1, \delta_2)$, we have $|f^{-1}(t)-f^{-1}(y)|=|f^{-1}(t)-x|<\epsilon'\le \epsilon,$ if $|t-y|<\delta.$ So $f^{-1}$ is continuous.

Note that $f$ doesn't even need to be continuous in the arguments above, and the extension of the proof to a strictly decreasing $f$, or $I=[a,b], [a,b),$ or $(a,b]$ is straightforward.

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