Question

could someone help me understand Topology of the Reals?!

Answer 1

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

Answer 2

i keep getting int and bd wrong

Answer 3

I wish I could help, dunno enough about topology yet

Answer 4

ok thank you though :)

Answer 5

I can help reason it out if you tell me definitions

Answer 6

Seems like interior is just stripping away the boundery.

Answer 7

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.

Answer 8

then x is called a boundary point of S.

Answer 9

You understand what neighborhood is?

Answer 10

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

Answer 11

and yes i understand what neighborhood is

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

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

Answer 16

I mean for a)

Answer 17

And it applies to c) and d) apparently

Answer 18

@manjuthottam make sense?

Answer 19

so the answer is focusing on rational numbers?

Answer 20

Okay, @manjuthottam do you know what continuity is?

Answer 21

is it just that the limit exists?

Answer 22

yeah the limit exists at the function

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

i'm guessing no?

Answer 28

at least its not that specific of a number

Answer 29

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.

Answer 30

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

Answer 31

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?

Answer 32

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.

Answer 33

In this way, neighborhoods are analogous to continuity.

Answer 34

hmm ok

Answer 35

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

Answer 36

yes it does

Answer 37

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

Answer 38

yes i see

Answer 39

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.

Answer 40

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

Answer 41

thus it is not in our set

Answer 42

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.

Answer 43

hmm ok that makes sense!

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

ok

Answer 52

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.

Answer 53

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

Answer 54

ok so even union of sets

Answer 55

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.

Answer 56

Yeah, it should work for unions of sets well too

Just not for intersections of sets.

Answer 57

so like e)

Answer 58

automatically the int is empty set

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

yeah

Answer 63

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

Answer 64

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