## anonymous 3 years ago could someone help me understand Topology of the Reals?!

1. anonymous

attached is the solutions for some question relating to it. the first 5 pages deal with it

2. anonymous

i keep getting int and bd wrong

3. anonymous

I wish I could help, dunno enough about topology yet

4. anonymous

ok thank you though :)

5. anonymous

I can help reason it out if you tell me definitions

6. anonymous

Seems like interior is just stripping away the boundery.

7. anonymous

Defn of int and bd: Let S be a subset of R. a point x in R is an interior point of S if there exists a neighborhood N of x such that N is a subset of S. if for every neighborhood N of x, N (junction) S does not = empty set. and N junction of (R\S) does not equal empty set.

8. anonymous

then x is called a boundary point of S.

9. anonymous

You understand what neighborhood is?

10. anonymous

well for first problem a) why is the interior empty set while for b) it is the whole set with open parentheses

11. anonymous

and yes i understand what neighborhood is

12. anonymous

what made 1/n interior pt be special to be empty while the union of two sets interior pt is the complete whole thing

13. anonymous

i would have guessed the ans for b) to be 3 or something

14. anonymous

Well, the problem is that you have: $1/1,1/2,1/3,1/4$Which is just rational a subset of rational numbers. The rational numbers are no continuous, so neither would be this subset. You can find an irrational number between any two elements, for example. Thus they are all boundary points.

15. anonymous

Despite not understanding neighborhood rigorously, that is my reasoning for b)

16. anonymous

I mean for a)

17. anonymous

And it applies to c) and d) apparently

18. anonymous

@manjuthottam make sense?

19. anonymous

so the answer is focusing on rational numbers?

20. anonymous

Okay, @manjuthottam do you know what continuity is?

21. anonymous

is it just that the limit exists?

22. anonymous

yeah the limit exists at the function

23. anonymous

But actually, let's talk about neighborhoods a bit ok?

24. anonymous

ok so neighborhood is x in R. while there is an epsilon greater than zero where |x-y|< epsilon?

25. anonymous

Yeah, notice how that looks kinda like a limit.... Anyway...

26. anonymous

Consider 1/2, 1/3, and 1/4 Do they have a neighborhood?

27. anonymous

i'm guessing no?

28. anonymous

at least its not that specific of a number

29. anonymous

If you think about it, they're equal to 12/24, 8/24, and 6/24 Since you don't have 10/24 or 4/24 in there, you can't really say they make up a neighborhood.

30. anonymous

So can you understand how these three points would be boundary points?

31. anonymous

so since there are values in between like 10/24 = (5/12) that are not part of the set,, there is no neighborhoood? ... because it is not a continuous set?

32. anonymous

Well, consider if we were to make a function: $f:\mathbb{R}\mapsto\{0,1\}$We let it be $$0$$ if $$x\in S$$ and $$1$$ if $$x\notin S$$ In this case, a neighborhood is just an interval for which $$f$$ is continuous.

33. anonymous

In this way, neighborhoods are analogous to continuity.

34. anonymous

hmm ok

35. anonymous

Does my analogy sound consistent to you understanding of what a neighborhood is?

36. anonymous

yes it does

37. anonymous

Okay so suppose we did have two points that were directly next to each other... they'd have to be $\frac{1}{n}, \frac{1}{n+1}$

38. anonymous

yes i see

39. anonymous

Notice how they are equal to: $\frac{2}{2n}, \frac{2}{2n+2}$And the thing is: $\frac{2}{2n+1}$is between these two points.

40. anonymous

And since the numerator is even and denominator is odd, we can't reduce this fraction anymore, so it can't be expressed as $\frac{1}{k},k\in\mathbb{N}$

41. anonymous

thus it is not in our set

42. anonymous

So no matter how big you make $$n, n+1$$, we can still find a number between them... you'll never have any continuity for our function $$f$$ so you'll never find a neighborhood.

43. anonymous

hmm ok that makes sense!

44. anonymous

This is sort of a proof by contradiction as to why there can't be any neighborhoods.

45. anonymous

ok, so then based on the defn of int and bd, we have to first see about continuity and neighborhood

46. anonymous

if it is continuous then we can see that int and bd are not an empty set

47. anonymous

The continuity thing was more to help me understand neighborhoods than it was to help you. =)

48. anonymous

=D kk but it helped me learn too! lol!

49. anonymous

I think what you need to understand is that rational numbers do NOT make any neighborhoods.

50. anonymous

you'll always be able to find some irrational number between two rational numbers.

51. anonymous

ok

52. anonymous

Thus when they give you a set involving rational numbers or natural numbers, etc. Remember that it must all be boundaries, no neighborhoods and no interior sets.

53. anonymous

If you get something in interval notation, just remember that changing [ ] to ( ) should give you interior set.

54. anonymous

ok so even union of sets

55. anonymous

There is a trick question in there where they use $$\cap$$ which is the set intersection... in those cases it has to be points in BOTH sets.

56. anonymous

Yeah, it should work for unions of sets well too Just not for intersections of sets.

57. anonymous

so like e)

58. anonymous

automatically the int is empty set

59. anonymous

For example: $[1,4]\cap [2,6]$This gives us: $[2,4]$this you can change them... $int([2,4]) = (2,4)\quad bd([2,4]) = \{2,4\}$

60. anonymous

So for intersections... just compute the intersection. It shouldn't be too bad.

61. anonymous

oh ok, so in sets that have a intersection we apply that method you mentioned but if the intersection is just one point like in e) then int is empty set because there is no neighborhood

62. anonymous

yeah

63. anonymous

You'd have to give me some $$\epsilon>0$$ but clearly $$2+\epsilon$$ is NOT in the set $$\{2\}$$

64. anonymous

kk thank you so much for helping me !!!!