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

[bohregard]

Homework Statement

Show that if f: [a,b]→$Re$ is increasing and the range of f is a bounded interval then f is continuous.

Homework Equations

N/A

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 {x_n} such that lim$_{n\rightarrow b}$ x_n = 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.

 

[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.

 

[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.

 

[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.

 

[jgens]

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

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.

 

[Robert1986]

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.

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.