## anonymous 3 years ago I need some help about topology homework.. anyone..

1. anonymous

2. anonymous

@helder_edwin

3. 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 $$|x|<\delta\implies \sin(\frac{1}{x})<\frac{1}{2}$$

4. 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 $$\sin(\frac{1}{x})=1$$

5. anonymous

yes I believe we can choose epsilon, since it is negation..

6. anonymous

thanks it makes sense to me..

7. anonymous

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. anonymous

@satellite73 hi, satellite73, can you check my comment here

9. 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

10. anonymous

definition of $\lim_{x\to a}f(x)=L$ $$\forall \epsilon> 0$$, $$\exists \delta$$ such that $$|x-a|\implies |f(x)-L|<\epsilon$$

11. anonymous

yes..

12. anonymous

it should be I guess $|x-a|\implies |f(x)-L|\ge \epsilon$

13. anonymous

negation is $$\exists \epsilon >0$$ such that $$\forall \delta$$, $$\exists x$$ such that $$|x-a|<\delta$$ and $$|f(x)-L|>\epsilon$$

14. anonymous

yes this is what I know..

15. 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

16. anonymous

in your example, you have to show that $$\exists \epsilon>0$$ so that $$\forall \delta$$ $$\exists x$$ with $$|x|<\delta$$ and $$|\sin(\frac{1}{x})|>\epsilon$$

17. anonymous

yes ..

18. anonymous

but it is gonna be contradiction right, and we are gonna say f(x) is not cont. at x=0..

19. 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 $$\sin(\frac{(2k+1)\pi}{2})=1>\frac{1}{2}$$

20. 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

21. 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

22. anonymous

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

23. anonymous

not greater than.

24. anonymous

i should have said $$\forall \delta>0$$ $$\exists k$$ such that $$\frac{1}{(2k+1)\pi}<\delta$$

25. anonymous

it makes more sense now..

26. anonymous

be careful here, you cannot write $\sin(\frac{1}{x})=0$ that makes no sense

27. 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

28. anonymous

thanks.. it must be grater than 0

29. 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

30. anonymous

yeah, I thought something like that.. how about 5)ii)

31. 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..