"A function which takes on infinite values throughout an interval is at least once continuous throughout a sub interval of that interval"
Prove or disprove the above statement<<<
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2Why should I??? – Gerry Myerson Aug 23 '13 at 12:57
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Do you know any examples of everywhere discontinuous functions? – Anthony Carapetis Aug 23 '13 at 12:58
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I should have been polite...I was trying to do so but I couldn't, please can you do so for me?? – Tom Lynd Aug 23 '13 at 12:59
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f(x)=x if x is irrational otherwise 0. the function is discontinuous everywhere except 0 – Tom Lynd Aug 23 '13 at 13:02
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1What does "takes on infinite values throughout an interval" mean? – hmakholm left over Monica Aug 23 '13 at 13:03
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I second Henning's question. As written, the question is too imprecise to answer definitively. – Cameron Buie Aug 23 '13 at 13:05
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e.g f(x)=sin x if x>0.4 otherwise 0 – Tom Lynd Aug 23 '13 at 13:07
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@TomLynd: Is that an attempt to explain what "takes on infinite values throughout an interval" means? It doesn't do that. – hmakholm left over Monica Aug 23 '13 at 13:10
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@Henning Makholm yes...trignometric functions are continous so how come sinx doesn't have infinite values or else you misunderstand me> – Tom Lynd Aug 23 '13 at 13:15
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1@Tom: I don't understand you at all. What does "have infinite values" mean? I don't want an example, but an understandable definition of this phrase. – hmakholm left over Monica Aug 23 '13 at 13:15
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@ Henning Makholm sin x will have infinite values throughout the interval(-1,1) that's how its graph is unbroken curve>..(the completeness property of real number system) – Tom Lynd Aug 23 '13 at 13:20
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2Tom, you're still not telling us what you mean by "have infinite values". Maybe you mean, "takes on infinitely many different values". Or maybe you mean, "take on every real value between its maximum and its minimum". Or maybe you mean, "only takes on the value infinity". Or maybe you mean something else. Please try again. – Gerry Myerson Aug 23 '13 at 13:26
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I mean infinite real numbers between its max and min – Tom Lynd Aug 23 '13 at 13:36
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There's no such thing as an infinite real number... – hmakholm left over Monica Aug 23 '13 at 14:55
1 Answers
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[EDIT: OP has deleted the comment which referred to the function taking on every real value between its minimum and its maximum, so perhaps this answer does not speak to OP's concerns. I leave it as a simple example of a nowhere-continuous bijection of a closed interval with itself. I also take the opportunity to correct a typo.]
From the comments, it seems the question is, if a function on an interval takes on every real value between its minimum and its maximum, must it be continuous somewhere on the interval?
A simple counterexample goes as follows:
Define $f:[0,1]\to[0,1]$ by $f(x)=x$ if $x$ is rational; $f(x)=x+(1/2)$ is $x$ is irrational and less than $1/2$; $f(x)=x-(1/2)$ if $x$ is irrational and exceeds $1/2$.
Gerry Myerson
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No,the function takes just infinite values between its max and min,not every possible real number> – Tom Lynd Aug 25 '13 at 07:25
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For the 18th and last time, the function does NOT take on "infinite values". It MAY take on infinitely many values, but every blessed one of those values is finite. AND my example takes on infinitely many values, doesn't it? But it still isn't continuous anywhere, is it? So, it answers your question, doesn't it? – Gerry Myerson Aug 25 '13 at 07:58