Baby Rudin continuity problem question

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Homework Help Overview

The discussion revolves around a problem from Baby Rudin concerning the continuity of a real function defined on R. The specific question is whether the condition that the limit of the difference quotient approaches zero implies that the function is continuous.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reason that the condition does not imply continuity, suggesting it only indicates the absence of simple discontinuities. Other participants question the definition of "simple" discontinuities and provide examples to illustrate their points.

Discussion Status

The discussion is active, with participants exploring different interpretations of discontinuities and clarifying the implications of the given limit condition. There is no explicit consensus, but participants are engaging with each other's ideas and refining their understanding.

Contextual Notes

Participants are discussing the implications of a specific mathematical condition without providing a definitive resolution. The nature of continuity and discontinuities is under examination, with some assumptions about definitions being questioned.

genericusrnme
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Sup guys, I was just going over my Baby Rudin and I came across a problem that I don't really know how to get started on.

Suppose f is a real function defined on R that satisfies, for all x [itex]Limit_{n\ \rightarrow \ 0} (f(x+n)-f(x-n)) = 0[/itex], does this imply f is continuous?

My first thoughts are that no, it doesn't imply f is continuous, it just implies that f doesn't have any simple discontinuities since [itex]f(x_+) = f_(x_-)[/itex]. I don't know how I can go about showing this though..

Could anyone nudge me in the right direction?

Thanks in advance!
 
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hi genericusrnme! :smile:
genericusrnme said:
… it just implies that f doesn't have any simple discontinuities …

does it ? :wink:
 
Well, what, exactly, do you mean by a "simple" discontinuity? If f(x)= 1 for all x except 0 and f(0)= 0, that looks like a pretty simple discontinuity to me!
 
tiny-tim said:
hi genericusrnme! :smile:


does it ? :wink:

Ah yes, you're completely right
f(x+) = f(x-) but f(x+) isn't necessarily equal to f(x)

HallsofIvy said:
Well, what, exactly, do you mean by a "simple" discontinuity? If f(x)= 1 for all x except 0 and f(0)= 0, that looks like a pretty simple discontinuity to me!

Yep, I just got that

Thanks guys!
 

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