Simple analysis continuity problem

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Homework Help Overview

The discussion revolves around a continuity problem involving a real function f that is continuous at a point a, with the condition that f(a) is less than a certain value M. The goal is to establish the existence of an open interval around a where f(x) remains less than M.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of continuity, particularly through the epsilon-delta definition, and how it relates to bounding the function f(x) within a certain distance of f(a). Questions arise regarding the application of this definition and its guarantees.

Discussion Status

The discussion is active, with some participants expressing clarity on the problem while others seek further understanding of the epsilon-delta definition and its role in the proof. There is an indication of productive exploration of the concepts involved.

Contextual Notes

One participant notes their inexperience with the topic, which may influence the depth of their contributions and understanding.

economist1985
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Homework Statement


If f is a real function which is continuous at a element R and if f(a)<M for some M element of R, prove that there is an open interval I containing a such that f(x)<M for all x element of I.


Homework Equations


Extreme value theorem, intermediate value theorem, definition of continuity


The Attempt at a Solution


I have no idea how to solve this.
 
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Draw out the scenario. You have a continuous function with some point a on the x-axis, f(a) on the y-axis. Draw a horizontal line through the point f(a). We can think of M as a horizontal line somewhere above f(a). Let D be the distance between f(a) and M. If we can make f(x) within D of f(a), then M will be a bound for f(x) as well. But this is guaranteed by the continuity of f (epsilon-delta definition).
 
Thanks for the explanation. The problem is clearer now to me. What I still do not get is how the epsilon delta definition guarantees it. ?? Sorry if I can't follow through... I'm new to this.
 
Oh I think I got it. Thanks
 

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