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? 
 A: 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 $M$.
You can then use the fact that continuity on a compact set implies uniform continuity, and you're done. This theorem, btw, isn't hard to prove either (and the proof shows how powerful the compactness property can be). The proof goes like this:
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 $$
  |f(x)-f(y)| < \epsilon \text{.}
$$ 
A: Here is a solution which uses only the intermediate value theorem for continuous functions.
Define $m = \inf_{t \in (x, y)} f(t)$ and $M = \sup_{t \in (x, y)} f(t)$. We can assume that $m < M$ because otherwise $f$ is constant.
Now let  $\epsilon > 0$ be given. The idea is to partition the range $[m, M]$ into intervals of length $< \epsilon/2$, then partition the domain $(x, y)$ into “corresponding” intervals, and then define $\delta$ as the smallest length of the domain intervals.
So we start by choosing a positive integer $n$ such that $\frac{M-m}n < \frac \epsilon 2$. Using the intermediate value theorem, we can find a partition
$$
 x = t_0 < t_1 < t_2 < \cdots < t_{n-1} < t_n = y
$$
such that
$$
f(t_j) = \frac{n-j}{n}m + \frac{j}{n}M 
$$
for $ 1 \le j \le n-1$. Note that $0 \le f(t_j) - f(t_{j-1}) < \frac \epsilon 2$ for $1 \le j \le n$. Finally define $\delta = \min_{1 \le j \le n}(t_j - t_{j-1})$.
Now, if $x < t < t^* < y$ and $t^* - t < \delta$ then only two cases are possible:

*

*$t$ and $t^*$ are in the same interval $[t_{j-1}, t_j]$. Then $0 \le f(t^*) - f(t) \le f(t_j) - f(t_{j-1}) < \frac \epsilon 2$.


*$t$ and $t^*$ are  in adjacent intervals $[t_{j-1}, t_j]$ and $[t_{j}, t_{j+1}]$. Then $0 \le f(t^*) - f(t) \le f(t_{j+1}) - f(t_{j-1}) < \epsilon$.
This shows that $|f(t^*) - f(t)| < \epsilon$ for all $t, t^* \in (x, y)$  with $|t-t^*| < \delta$.
Remark: The same proof works with small modifications if the domain is an unbounded interval $(x, \infty)$, $(-\infty, y)$, or $\Bbb R$.
A: *

*Define a new function g on [a b]

Our first task is to extend f to a continuous function g defined on [a b].
Let's call the one-sided limits of f(x)  as  $l_1$ and $l_2$ when x approaches a and b from the right or left respectively.  More specifically,
lim f(x) =l1 as x->a+, and lim f(x) =l2 as x->b−.
Note that since f is increasing and bounded on (a b), both limits exist
and are finite.
Define g: [a b]--> R as follows:
g(x)= f(x) for all x ε (a b),
g(a) = l1 and g(b) = l2.
Hence g is an extension of f to [a b] and it can be shown that g
is continuous.


*Show that g is uniformly continuous on [a b].

Since g is continuous on [a b]. Hence it is also uniformly continuous on [a b].
(see for example Theorem 4.19 in Walter Rudin "Principles of
Mathematical Analysis", 2nd ed. p 78)
Hence f, being the restriction of g to (a b), is also uniformly
continuous on (a b).
