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

  1. Apr 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that if f: [a,b]→[itex]Re[/itex] 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]→[itex]Re[/itex] increasing and the rg(f) be a bounded interval. Since f is increasing, there exists x,y[itex]\in[/itex][a,b] such that f(a)≤f(x)<f(y)≤f(b). Further, since [a,b] is a bounded interval, S={x|x[itex]\in[/itex][a,b]}, there's a sequence {xn} such that lim[itex]_{n\rightarrow b}[/itex] 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. jcsd
  3. Apr 2, 2012 #2

    jgens

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    Gold Member

    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 [itex]x \in (a,b)[/itex] and suppose [itex]0 < \varepsilon[/itex]. Define [itex]\delta_1 = \sup\{y \in [x,b]: 0 \leq f(y)-f(x) < \varepsilon\}[/itex] and [itex]\delta_2 = \inf\{y \in [a,x]: 0 \leq f(x)-f(y) < \varepsilon\}[/itex]. You can show that both [itex]0 < \delta_1,\delta_2[/itex] using the fact that [itex]f([a,b])[/itex] is an interval. Then set [itex]\delta = \min\{\delta_1,\delta_2\}[/itex] and check that the [itex]\varepsilon,\delta[/itex] criterion holds. The argument when [itex]x = a[/itex] or [itex]x = b[/itex] 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.
     
  4. Apr 2, 2012 #3
    This doesn't seem to be true to me.

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

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


    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
  5. Apr 2, 2012 #4
    Yes, I am quite sure that this is not true.

    I think that it should be something like:

    [itex]f[/itex] 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 [itex]\{y\in[0,b]:0\leq f(y) - f(0) < 1/2 \}[/itex] is empty.
     
  6. Apr 2, 2012 #5

    jgens

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    Gold Member

    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.
     
  7. Apr 2, 2012 #6
    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.
     
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