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cinar Group Title

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

  • one year ago
  • one year ago

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  1. cinar Group Title
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    • one year ago
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  2. cinar Group Title
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    @helder_edwin

    • one year ago
  3. satellite73 Group Title
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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 \(|x|<\delta\implies \sin(\frac{1}{x})<\frac{1}{2}\)

    • one year ago
  4. satellite73 Group Title
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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 \(\sin(\frac{1}{x})=1\)

    • one year ago
  5. cinar Group Title
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    yes I believe we can choose epsilon, since it is negation..

    • one year ago
  6. cinar Group Title
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    thanks it makes sense to me..

    • one year ago
  7. cinar Group Title
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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..

    • one year ago
  8. cinar Group Title
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    @satellite73 hi, satellite73, can you check my comment here

    • one year ago
  9. satellite73 Group Title
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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

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

    • one year ago
  11. cinar Group Title
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    yes..

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

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

    • one year ago
  14. cinar Group Title
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    yes this is what I know..

    • one year ago
  15. satellite73 Group Title
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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

    • one year ago
  16. satellite73 Group Title
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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 \(|\sin(\frac{1}{x})|>\epsilon\)

    • one year ago
  17. cinar Group Title
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    yes ..

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

    • one year ago
  19. satellite73 Group Title
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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 \(\sin(\frac{(2k+1)\pi}{2})=1>\frac{1}{2}\)

    • one year ago
  20. satellite73 Group Title
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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

    • one year ago
  21. satellite73 Group Title
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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

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

    • one year ago
  23. cinar Group Title
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    not greater than.

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

    • one year ago
  25. cinar Group Title
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    it makes more sense now..

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

    • one year ago
  27. satellite73 Group Title
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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

    • one year ago
  28. cinar Group Title
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    thanks.. it must be grater than 0

    • one year ago
  29. satellite73 Group Title
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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

    • one year ago
  30. cinar Group Title
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    yeah, I thought something like that.. how about 5)ii)

    • one year ago
  31. cinar Group Title
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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..

    • one year ago
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