# Show f is continuous if the range of f is a bounded interval

1. Apr 2, 2012

### bohregard

1. The problem statement, all variables and given/known data
Show that if f: [a,b]→$Re$ is increasing and the range of f is a bounded interval then f is continuous.

2. Relevant equations
N/A

3. The attempt at a solution
I have no idea where to start, but I decided to start with a couple of things.

Proof: Let f: [a,b]→$Re$ increasing and the rg(f) be a bounded interval. Since f is increasing, there exists x,y$\in$[a,b] such that f(a)≤f(x)<f(y)≤f(b). Further, since [a,b] is a bounded interval, S={x|x$\in$[a,b]}, there's a sequence {xn} such that lim$_{n\rightarrow b}$ xn = f(b). Similarly for lim = f(a).

Now I'm not really sure where to go, or if I was even headed in the right direction.

2. Apr 2, 2012

### jgens

If you have the open set formulation of continuity, then this problem is pretty simple. If you do not have that formulation of continuity, then fix $x \in (a,b)$ and suppose $0 < \varepsilon$. Define $\delta_1 = \sup\{y \in [x,b]: 0 \leq f(y)-f(x) < \varepsilon\}$ and $\delta_2 = \inf\{y \in [a,x]: 0 \leq f(x)-f(y) < \varepsilon\}$. You can show that both $0 < \delta_1,\delta_2$ using the fact that $f([a,b])$ is an interval. Then set $\delta = \min\{\delta_1,\delta_2\}$ and check that the $\varepsilon,\delta$ criterion holds. The argument when $x = a$ or $x = b$ is similar.

I think most people would probably do this proof by contradiction and I can walk you through that if you are more comfortable with that sort of proof.

3. Apr 2, 2012

### Robert1986

This doesn't seem to be true to me.

For example, let
$$f(x) = \begin{cases} \arctan(x), & \text{if } x\text{ \le 0} \\ \arctan(x) + 1, & \text{if } x\text{ > 0} \end{cases}$$

This is increasing on any interval $[a,b]$, isn't it? And the range of $f$ is $(-\pi/2, \pi/2 +1)$, right? But it has a jump at $0$.

BTW, if I'm wrong about this let me know. But, also, can someone tell me how to get the less than or equal symbol? I tried \le and \leq both to no avail.

Last edited: Apr 2, 2012
4. Apr 2, 2012

### Robert1986

Yes, I am quite sure that this is not true.

I think that it should be something like:

$f$ is continuous except for a finite number of jumps.

Then jgens solution could be applied to all non-jumps.

Using the example I did in my last post, the set $\{y\in[0,b]:0\leq f(y) - f(0) < 1/2 \}$ is empty.

5. Apr 2, 2012

### jgens

I interpreted range(f) as image(f) since that is a fairly common convention (I think analysts use this terminology). With this convention the result is certainly true.

I agree that the use of range is ambiguous and that is why I do not use it. I prefer to use the terms image and codomain since the meanings of these terms is more uniform. Unfortunately not everyone agrees.

6. Apr 2, 2012

### Robert1986

Yeah, that makes sense, I'd say you're right about that. I interpreted it as "range(f) is a bounded subset of R". I think I interpreted it incorrectly.