## anonymous one year ago here

1. anonymous

Let R be endowed with the usual standard topology. Consider Y = [-1,1] as a subspace of R. Which one of the following sets is closed in Y

2. anonymous

@zzr0ck3r

3. zzr0ck3r

?

4. anonymous

{x: $\frac{1}{2}\ <$ |x| < 1}

5. zzr0ck3r

not closed

6. anonymous

is that open?

7. anonymous

{x:$\frac{1}{2}$ <$|x|\leqslant$1}

8. zzr0ck3r

yes, since it is open in R and and it equals the intersection with the subspace.

9. zzr0ck3r

its open

10. anonymous

${x: \frac{1}{2}\leqslant|x| < 1}$

11. zzr0ck3r

neither open nor closed

12. anonymous

${x: \frac{1}{2}\leqslant|x| < 1}$

13. zzr0ck3r

thats the same. please use parentheses. bbl

14. anonymous

${x: \frac{1}{2} < |x|\leqslant1}$

15. anonymous

${x: \frac{1}{2}\leqslant|x|\leqslant1}$

16. anonymous

i presume this is close

17. zzr0ck3r

yep

18. zzr0ck3r

closed sets in a subspace S, are of the form $$A\cap S$$ where $$A$$ is open in the parent topology.

19. zzr0ck3r

ok walking out the door

20. anonymous

and which is open in Y with same options ?

21. anonymous

is it ${x: \frac{1}{2}\ < |x| < 1}$

22. anonymous

2 With the standard topology on R,which one of the sets in question (1) above is open in R? with same options given above?

23. anonymous

With the standard topology on R,which one of the sets in question (1) above is closed in R? with same options given above

24. zzr0ck3r

hi

25. anonymous

sir, i am here @zzr0ck3r

26. zzr0ck3r

ok

27. anonymous

so, can you help with the asked question sir?

28. zzr0ck3r

$$(-1, -\dfrac{1}{2})\cup (\dfrac{1}{2}, 1)$$ is open $$(-1, -\dfrac{1}{2}]\cup [\dfrac{1}{2}, 1)$$ is not open $$[-1, -\dfrac{1}{2})\cup (\dfrac{1}{2}, 1]$$ is not open $$[-1, -\dfrac{1}{2}]\cup [\dfrac{1}{2}, 1]$$ is not open

29. anonymous

so which means that $x:1/2 <|x|<1$ is open in the usual standard toplogy

30. zzr0ck3r

correct

31. anonymous

what about in the standard topology on R

32. anonymous

2 With the standard topology on R,which one of the sets in question (1) above is open in R with same options given above?

33. anonymous

you here sir?

34. anonymous

@zzr0ck3r

35. anonymous

hello sir, @zzr0ck3r

36. anonymous

let me post the question in full . please don't get mad at me

37. anonymous

With the standard topology on R,which one of the sets is open in R?

38. anonymous

${x: \frac{1}{2}\ < |x| < 1}$ ${x: \frac{1}{2} < |x|\leqslant1}$ ${x: \frac{1}{2}\leqslant|x| < 1}$ ${x: \frac{1}{2}\leqslant|x|\leqslant1}$

39. anonymous

is the answer still the first here sir?

40. anonymous

i guess from what you thought, A is the only open set while B,C, D are close sets

41. anonymous

am i correct?

42. anonymous

another question

43. anonymous

A topological space is called a Hausdorff space, if for each x, y of distinct points of X, there exists nbds $U_{x}$ and $U_{y}$ of x and y respectively that are disjoint. This implies X is Hausdorff wiith these properties except one.

44. anonymous

A) if $\forall x, y \epsilon\mathbb X; x \neg y$

45. anonymous

B)$There \exists U_x \epsilon N(x). U_y \epsilon N(y)$

46. anonymous

C) $U_x \bigcap U-y = \phi$

47. anonymous

D)$\forall x, y \epsilon X, x \bigcap y = 0$

48. zzr0ck3r

what is the question?

49. anonymous

should i post it in anew question?

50. zzr0ck3r

yes