Increasing, bounded and continuous is uniformly continuous

Let $f:(x,y)\to \mathbb{R}$ be increasing, bounded and continuous on $(x,y)$. Prove that $f$ is uniformly continuous on $(x,y)$.

Since it is bounded then there exists an $M\in \mathbb{R}$ such that $|f(x,y)|<M$ and since it is continuous then $$\forall \epsilon > 0, \forall x \in X, \exists \delta > 0 : |x - y| < \delta \implies |f(x) - f(y)| < \epsilon.$$ To show it is uniformly continuous I must show: $$\text{there exists} \ \epsilon >0 \ \forall \ \delta \ \exists \ (x_0,x)\in I:\{|x -x_0|<\delta \implies |f(x)-f(x_0)|\le \epsilon\}$$

but I am not sure how I can show that?

• I've added a proof of the Heine-Cantor theorem, i.e. that continuity on a compact set implies uniform continuity to my answer. I hope I've managed to highligh how compactness is the deciding property here, by allowing you to restrict your attention to only finitly many subsets. – fgp Apr 30 '13 at 12:03

Since $f$ is continuous on $(a,b)$, bounded and increasing, there's a unique continous extension of $f$ to $[a,b]$. This works because both limits $f(b) := \lim_{x\to b-}$ and $f(a) = \lim_{x\to a+}$ and are guaranteed to exist since every bounded and increasing (respectively bounded a decreasing) sequence converges. To prove this, simply observe that for a increasing and bounded sequence, all $x_m$ with $m > n$ have to lie within $[x_n,M]$ where $M=\sup_n x_n$ is the upper bound. Add to that the fact that by the very definition of $\sup$, there are $x_n$ arbitrarily close to $M$.
First, recall the if $f$ is continuous then the preimage of an open set, and in particular of an open interval, is open. Thus, for $x \in [a,b]$ all the sets $$C_x := f^{-1}\left(\left(f(x)-\frac{\epsilon}{2},f(x)+\frac{\epsilon}{2}\right)\right)$$ are open. The crucial property of these $C_x$ is that for all $y \in C_x$ you have $|f(y)-f(x)| < \frac{\epsilon}{2}$ and thus $$|f(u) - f(v)| = |(f(u) - f(x)) - (f(v)-f(x))| \leq \underbrace{|f(u)-f(x)|}_{<\frac{\epsilon}{2}} + \underbrace{|f(v)-f(x)|}_{<\frac{\epsilon}{2}} < \epsilon \text{ for all } u,v \in C_x$$ Now recall that an open set contains an open interval around each of its points. Each $B_x$ thus contains an open interval around $x$, and you may wlog assume that its symmetric around $x$ (just make it smaller if it isn't). Thus, there are $$\delta_x > 0 \textrm{ such that } B_x := (x-\frac{\delta_x}{2},x+\frac{\delta_x}{2}) \subset (x-\delta_x,x+\delta_x) \subset C_x$$ Note how we made $B_x$ artifically smaller than seems necessary, that will simplify the last stage of the proof. Since $B_x$ contains $x$, the $B_x$ form an open cover of $[a,b]$, i.e. $$\bigcup_{x\in[a,b]} B_x \supset [a,b] \text{.}$$ Now we invoke compactness. Behold! Since $[a,b]$ is compact, every covering with open sets contains a finite covering. We can thus pick finitely many $x_i \in [a,b]$ such that we still have $$\bigcup_{1\leq i \leq n} B_{x_i} \supset [a,b] \text{.}$$ We're nearly there, all that remains are a few applications of the triangle inequality. Since we're only dealing with finitly many $x_i$ now, we can find the minimum of all their $\delta_{x_i}$. Like in the definition of the $B_x$, we leave ourselves a bit of space to maneuver later, and actually set $$\delta := \min_{1\leq i \leq n} \frac{\delta_{x_i}}{2} \text{.}$$
Now pick arbitrary $u,v \in [a,b]$ with $|u-v| < \delta$. Since our $B_{x_1},\ldots,B_{x_n}$ form a cover of $[a,b]$, there's an $i \in {1,\ldots,n}$ with $u \in B_{x_i}$, and thus $|u-x_i| < \frac{\delta_{x_i}}{2}$. Having been conservative in the definition of $B_x$ and $\delta$ pays off, because we get $$|v-x_i| = |v-((x_i-u)+u)| = |(v-u)-(x_i-u)| < \underbrace{|u-v|}_{<\delta\leq\frac{\delta_{x_i}}{2}} + \underbrace{|x_i-u|}_{<\frac{\delta_{x_i}}{2}} < \delta_{x_i} \text{.}$$ This doesn't imply $y \in B_{x_i}$ (the distance would have to be $\frac{\delta_{x_i}}{2}$ for that), but it does imply $y \in C_{x_i}$!. We thus have $x \in B_{x_i} \subset C_{x_i}$ and $y \in C_{x_i}$, and by definition of $C_x$ (see the remark about the crucial property of $C_x$ above) thus $$|f(x)-f(y)| < \epsilon \text{.}$$
• In the first paragraph of your answer, why is it sufficient to only consider monotonically increasing sequences $(x_{n})_{n=0}^{\infty}$ converging to $b$ (say)? Is it possible to prove that $\lim_{n\rightarrow\infty}f(x_{n})$ exits for any (not necessarily monotonically increasing) sequence $(x_{n})_{n=0}^{\infty}\rightarrow b$? – Karthik Kannan Sep 23 '19 at 18:28
• I guess we could prove it as follows: Let $(x_{n})_{n=0}^{\infty}\rightarrow b$ and let $L = \sup_{n\in\mathbb{N}}f(x_{n})<\infty$ since $f$ is bounded. So using the monotonicity of $f$, for every $\epsilon >0$ there exist $N_{\epsilon}$ such that $L-\epsilon < f(x_{n})\leq L$ for all $n\geq N_{\epsilon}$. Then we conclude that $\lim_{n\rightarrow\infty}f(x_{n}) = L$ by taking $\limsup_{n\rightarrow\infty}$ and $\liminf_{n\rightarrow\infty}$ on these inequalities. – Karthik Kannan Sep 23 '19 at 19:13
• The above may be slightly incorrect. Let $(x_{n})_{n=0}^{\infty}$ be any sequence in $(a, b)$ coverging to $b$ and let $L = \sup_{x\in (a, b)}f(x)$ so for every $\epsilon>0$ there exists $x_{0}\in(a, b)$ such that $L-\epsilon < x_{0}\leq L$. Since $(x_{n})_{n=0}^{\infty}\rightarrow b$ and $b-x_0>0$ there exists $N\in\mathbb{N}$ such that $b-(b-x_{0})\leq x_{n}$ for all $n\geq N$ and so $L-\epsilon< f(x_{0})\leq f(x_{n})\leq L$ for all $n\geq N$. The result then follows by taking $\limsup_{n\rightarrow\infty}$ and $\liminf_{n\rightarrow\infty}$. – Karthik Kannan Oct 14 '19 at 20:55