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If $f: \mathbb R \to \mathbb R$ is bounded, increasing and continuous. Does $f$ have to uniform continuous?


I know the answer is yes if $f$ has domain to be any open interval, say $(0,1)$. But I don't know how to prove this case. Any help is appreciated.

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marked as duplicate by user99914, mookid, Claude Leibovici, Yiyuan Lee, Etienne Mar 26 '14 at 8:10

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Yes, and here's an outline of how to prove it: Suppose that $L$ is the limit at $\infty$ and $M$ is the limit at $-\infty$ (do you see why both exist?). Choose an $\epsilon > 0$; then there's a compact set $K$ such that

$$f^{-1}\Big((M + \epsilon, L - \epsilon)\Big) \subseteq K$$

Now $f$ is uniformly continuous on $K$, and if $x, y$ are both large enough to be outside of $K$ (but close together), use the definition of $K$ to bound the quantity $|f(x) - f(y)|$.

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  • $\begingroup$ Thank you so much! I got it! $\endgroup$ – user112564 Mar 26 '14 at 4:37
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Because the function is bounded and monotonically we know that the limits of the function at the endpoints (WLOG 0 and 1) exist. Now define $F:[0,1]->\mathbb{R}$ by $F=f$ on the open interval and $F(0)=\lim_{x\to 0} f(x)$ and likewise for $F(1)$. Then $F(x)$ is continuous on a compact set and therefore uniformily continuous (just apply total boundedness), and therefore $f$ is too.

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