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Prove that a strictly increasing function $f:[a,b]\rightarrow\mathbb{R}$ which has the intermediate value property is continuous on $[a,b]$.

Let $x_0\in[a,b]$. As $f$ is strictly increasing, $$\sup_{a\leq x<x_0}f(x)=f(x_0^+)\leq f(x)\leq\inf_{x_0<x\leq b}f(x)=f(x_0^-)$$ To yield a contradiction, suppose that $f(x)<f(x_0^+)$. Let $\{x_n\}$ be a sequence of $(x_0,b]$ that converges to $x_0^+$such that $\lim\limits_{n\rightarrow\infty}f(x_n)=f(x_0^+)$. Since $f$ is strictly increasing, $$f(x_n)>f(x_0^+)>f(x_0)$$ Apply the intermediate value theorem, there exists a point $x'\in(x_0,x_n]$ such that $f(x')=f(x_0^+)$, then $$\inf_{x_0\leq x<x'}f(x)\geq\inf_{x_0<x\leq b}f(x)=f(x')$$ However, by strictly monotonicity of $f$, $\inf_{x_0\leq x<x'} f(x)<f(x')$ which is a contradiction.

Hence, $f(x)=f(x_0^+)$. Similarly, $f(x)=f(x_0^-)$,$f(a)=f(a^+)$, and $f(b)=f(b^-)$.

Can someone give a hint or suggestion to write a direct proof? Thanks

Simple
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1 Answers1

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Let $x$ be in the domain. Let $\epsilon>0$. If $f(x)-\epsilon$ and $f(x) + \epsilon$ are not in the range, then any $\delta$ will work. If at least one is, let $\delta$ be smaller than $\operatorname{min}(f^{-1}(f(x)+\epsilon) - x, f^{-1}(f(x) - \epsilon))$.

I think that works. Am I missing something?

Eric Auld
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  • you are implicitly using the fact that $f$ is onto some interval. This is a consequence of the intermediate value property but at the level of this discussion would require some reasoning. – Thomas Dec 06 '15 at 07:06
  • Is it not clear that if it is increasing and has IVP then it is onto $[f(a), f(b)]$? What other justification would you suggest? – Eric Auld Dec 06 '15 at 07:08
  • Just stating that fact would be sufficient and an improvement. – Thomas Dec 06 '15 at 08:41