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cinar

  • 2 years ago

I need some help about topology homework.. anyone..

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  1. cinar
    • 2 years ago
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  2. cinar
    • 2 years ago
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    @helder_edwin

  3. satellite73
    • 2 years ago
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    first one i can do negate the definition by picking \(\epsilon =\frac{1}{2}\) then show that there does not exist a \(\delta\) such that \(|x|<\delta\implies \sin(\frac{1}{x})<\frac{1}{2}\)

  4. satellite73
    • 2 years ago
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    that is because no matter how small \(\delta\) is there will be some \(k\) such that \[\frac{2}{(2k+1)\pi}<\delta\] and for that \(k\) and all other larger you will have \(\sin(\frac{1}{x})=1\)

  5. cinar
    • 2 years ago
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    yes I believe we can choose epsilon, since it is negation..

  6. cinar
    • 2 years ago
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    thanks it makes sense to me..

  7. cinar
    • 2 years ago
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    sin(1/x) must be greater than epsilon, and it is true, how about 1/(2kpi)<delta we have sin(1/x)=0 not greater than epsilon. can we use this..

  8. cinar
    • 2 years ago
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    @satellite73 hi, satellite73, can you check my comment here

  9. satellite73
    • 2 years ago
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    hi do you know exactly how to negate the definition of \[\lim_{x\to a}f(x)=L ?\]that is what you need to use

  10. satellite73
    • 2 years ago
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    definition of \[\lim_{x\to a}f(x)=L \] \(\forall \epsilon> 0\), \(\exists \delta \) such that \(|x-a|\implies |f(x)-L|<\epsilon\)

  11. cinar
    • 2 years ago
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    yes..

  12. cinar
    • 2 years ago
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    it should be I guess \[|x-a|\implies |f(x)-L|\ge \epsilon\]

  13. satellite73
    • 2 years ago
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    negation is \(\exists \epsilon >0\) such that \(\forall \delta\), \(\exists x\) such that \(|x-a|<\delta\) and \(|f(x)-L|>\epsilon\)

  14. cinar
    • 2 years ago
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    yes this is what I know..

  15. satellite73
    • 2 years ago
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    you need to use what i wrote above (took me a while) so the logic is this you exhibit a specific \(\epsilon\) and then show that no matter how small \(\delta\) is, then you can find some \(x\) for which \(|x-a|<\delta\) and at the same time for that \(x\) you have \(|f(x)-L|>\epsilon\) my suggestion was to pick \(\epsilon =\frac{1}{2}\) whatever you pick, it should be specific not general

  16. satellite73
    • 2 years ago
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    in your example, you have to show that \(\exists \epsilon>0\) so that \(\forall \delta\) \(\exists x\) with \(|x|<\delta\) and \(|\sin(\frac{1}{x})|>\epsilon\)

  17. cinar
    • 2 years ago
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    yes ..

  18. cinar
    • 2 years ago
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    but it is gonna be contradiction right, and we are gonna say f(x) is not cont. at x=0..

  19. satellite73
    • 2 years ago
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    since \(\sin(\frac{1}{x})=1\) infinitely often close to zero, you can show this pick \(\epsilon=\frac{1}{2}\) for example, then \(\forall \delta>0\) \(\exists \frac{1}{(2k+1)\pi}<\delta\) and so \(\sin(\frac{(2k+1)\pi}{2})=1>\frac{1}{2}\)

  20. satellite73
    • 2 years ago
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    you are negating the definition, but this is not a proof by contradiction, it is a direct proof, using the negation of the definition

  21. satellite73
    • 2 years ago
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    a proof by contradiction would be to assume it is continuous and then arrive at some contradiction, but you do not need to do this here. write down exactly what the negation of the definition is, then prove it directly

  22. cinar
    • 2 years ago
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    can I use my example to get contradiction..sin(1/x)=0 but greater than 1/2

  23. cinar
    • 2 years ago
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    not greater than.

  24. satellite73
    • 2 years ago
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    i should have said \(\forall \delta>0\) \(\exists k\) such that \(\frac{1}{(2k+1)\pi}<\delta\)

  25. cinar
    • 2 years ago
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    it makes more sense now..

  26. satellite73
    • 2 years ago
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    be careful here, you cannot write \[\sin(\frac{1}{x})=0\] that makes no sense

  27. satellite73
    • 2 years ago
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    again, i stress work the directly from the definition of " \(f\) is not continuous at \(a\)" i know it seems like a proof by contradiction, but it is not the definition of "\(f\) is not continuous at \(a\)" i wrote above

  28. cinar
    • 2 years ago
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    thanks.. it must be grater than 0

  29. satellite73
    • 2 years ago
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    question 2 looks annoying but i think you can do it by saying something like this for any two points you can join them by a straight line (you can even write down what the line would be) unless the line would contain those two points, in which case you can join them by a horizontal and vertical line that skips those points probably an easier way, not sure

  30. cinar
    • 2 years ago
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    yeah, I thought something like that.. how about 5)ii)

  31. cinar
    • 2 years ago
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    for 5) ii) can we say that since A is compact it is finite.. I know it is bounded and closed.. is it true that every closed and bounded set is finite..

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