anonymous
  • anonymous
I need some help about topology homework.. anyone..
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
@helder_edwin
anonymous
  • anonymous
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 \(|x|<\delta\implies \sin(\frac{1}{x})<\frac{1}{2}\)

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anonymous
  • anonymous
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 \(\sin(\frac{1}{x})=1\)
anonymous
  • anonymous
yes I believe we can choose epsilon, since it is negation..
anonymous
  • anonymous
thanks it makes sense to me..
anonymous
  • anonymous
sin(1/x) must be greater than epsilon, and it is true, how about 1/(2kpi)
anonymous
  • anonymous
@satellite73 hi, satellite73, can you check my comment here
anonymous
  • anonymous
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
anonymous
  • anonymous
definition of \[\lim_{x\to a}f(x)=L \] \(\forall \epsilon> 0\), \(\exists \delta \) such that \(|x-a|\implies |f(x)-L|<\epsilon\)
anonymous
  • anonymous
yes..
anonymous
  • anonymous
it should be I guess \[|x-a|\implies |f(x)-L|\ge \epsilon\]
anonymous
  • anonymous
negation is \(\exists \epsilon >0\) such that \(\forall \delta\), \(\exists x\) such that \(|x-a|<\delta\) and \(|f(x)-L|>\epsilon\)
anonymous
  • anonymous
yes this is what I know..
anonymous
  • anonymous
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
anonymous
  • anonymous
in your example, you have to show that \(\exists \epsilon>0\) so that \(\forall \delta\) \(\exists x\) with \(|x|<\delta\) and \(|\sin(\frac{1}{x})|>\epsilon\)
anonymous
  • anonymous
yes ..
anonymous
  • anonymous
but it is gonna be contradiction right, and we are gonna say f(x) is not cont. at x=0..
anonymous
  • anonymous
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 \(\sin(\frac{(2k+1)\pi}{2})=1>\frac{1}{2}\)
anonymous
  • anonymous
you are negating the definition, but this is not a proof by contradiction, it is a direct proof, using the negation of the definition
anonymous
  • anonymous
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
anonymous
  • anonymous
can I use my example to get contradiction..sin(1/x)=0 but greater than 1/2
anonymous
  • anonymous
not greater than.
anonymous
  • anonymous
i should have said \(\forall \delta>0\) \(\exists k\) such that \(\frac{1}{(2k+1)\pi}<\delta\)
anonymous
  • anonymous
it makes more sense now..
anonymous
  • anonymous
be careful here, you cannot write \[\sin(\frac{1}{x})=0\] that makes no sense
anonymous
  • anonymous
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
anonymous
  • anonymous
thanks.. it must be grater than 0
anonymous
  • anonymous
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
anonymous
  • anonymous
yeah, I thought something like that.. how about 5)ii)
anonymous
  • anonymous
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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