Proving Continuity of Monotone Functions on Interval Domains

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Homework Help Overview

The discussion revolves around proving the continuity of monotone functions defined on interval domains, specifically focusing on a function f:A-->R where the range of f is an interval. Participants explore the implications of monotonicity and the nature of discontinuities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of discontinuities in monotone functions, particularly jump discontinuities, and how the range being an interval affects continuity. There are attempts to formulate an analytical proof and questions about defining delta in relation to epsilon. Some participants suggest using the inverse function and its continuity as part of the proof.

Discussion Status

The discussion is active, with participants offering various approaches and questioning each other's reasoning. Some have provided hints and guidance on how to approach the proof, while others express confusion regarding the continuity of inverse functions and the implications of the problem statement.

Contextual Notes

There are indications of potential issues or ambiguities in the problem statement, particularly regarding the continuity of the inverse function and the specific examples being discussed. Participants are also considering the implications of the function's domain and range.

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Suppose f:A-->R is monotone (ACR: reals)
and suppose the range of f is an interval, show f is continuous on A.

By drawing a picture, I can see the conclusion. Since f is monotone, the only type of discontinuity it may have is a jump discontinuity. But since the range of f is an interval, this cannot happen.

I would like to have a more consistent proof (analytical proof), I mean, I know that since f is monotone, for each point a of A, f(a+) and f(a-) exist. Now f(a+)> or eq. to f(a) and f(a-)< or eq. to f(a). Now how would I show that f(a+)< or eq. to f(a) and f(a-)> or eq. to f(a) using the fact that range(f) is an interval? Hence I would be able to conclude that f(a-)= f(a+)=f(a), therefore, f continuous on A.
 
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Just take the direct approach. Pick an x and an epsilon e>0. You want to show that there is a delta d>0 such that |f(x)-f(y)|<e if |x-y|<d, right? Think about x=f^(-1)(f(x)), f^(-1)(f(x)+e) and f^(-1)(f(x)-e). Can you figure out how to define a d using those?
 
I would use d such that |f^(-1)(f(x)+e) - f^(-1)(f(x)-e)|> d. Is that correct? How do we use the fact that range of f is an interval?
I see that you're using f^(-1), in fact the problem has a subproblem that says to use the result to prove the continuity of inverse functions. How do we conclude that?
 
If there is a jump discontinuity, then the range isn't an interval. I use that the range is an interval to show that a point like f^(-1)(f(x)+e) even exists. You've probably drawn a picture of a function with a jump discontinuity, so you should be able to see how the proof breaks down. I would say d should be smaller than the maximum of |f^(-1)(f(x)+e)-x| and and |f^(-1)(f(x)-e)-x|. Do you see why? Try doing the subproblem about f^(-1) and see how far you get.
 
shouldn't d be smaller than the minimum of |f^(-1)(f(x)+e)-x| and |f^(-1)(f(x)-e)-x| instead?
 
Sure. Absolutely. I was just testing you. :)
 
oh, good, then, at least I understand what's going on...
 
That's what I was checking. Actually 'maximum' was just a typo...
 
I am trying for the f^-1 part, but I cannot get anywhere, can you give me another hint?
 
  • #10
What exactly is the f^(-1) part?
 
  • #11
the subproblem that says to use the result to prove the continuity of inverse functions.
 
  • #12
The inverse of a continuous function is not necessarily continuous even if it's monotone and the range is an interval. Can you state the full problem? Do you know something about the domain?
 
  • #13
that's it, that's how it is stated:

"Suppose f:A-->R is monotone (ACR: reals)
and suppose the range of f is an interval, show f is continuous on A. Use this result to prove the continuity of inverse functions".
 
  • #14
Let f(x)=x for x in [0,1) and f(x)=x-1 for x in [2,3]. f is continuous. The domain is [0,1)U[2,3] which is a subset of the reals. The range is [0,2]. It's monotone. The inverse has a jump discontinuity at 1. Do you see my problem?
 
  • #15
yeah, I guess, there is a problem or maybe something missing in the question, anyways, thanks for your help.
 

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