A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
could someone help me understand Topology of the Reals?!
anonymous
 3 years ago
could someone help me understand Topology of the Reals?!

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0attached is the solutions for some question relating to it. the first 5 pages deal with it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i keep getting int and bd wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I wish I could help, dunno enough about topology yet

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok thank you though :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can help reason it out if you tell me definitions

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Seems like interior is just stripping away the boundery.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Defn 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then x is called a boundary point of S.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You understand what neighborhood is?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well for first problem a) why is the interior empty set while for b) it is the whole set with open parentheses

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and yes i understand what neighborhood is

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what made 1/n interior pt be special to be empty while the union of two sets interior pt is the complete whole thing

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i would have guessed the ans for b) to be 3 or something

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Despite not understanding neighborhood rigorously, that is my reasoning for b)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And it applies to c) and d) apparently

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@manjuthottam make sense?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the answer is focusing on rational numbers?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, @manjuthottam do you know what continuity is?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is it just that the limit exists?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah the limit exists at the function

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But actually, let's talk about neighborhoods a bit ok?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so neighborhood is x in R. while there is an epsilon greater than zero where xy< epsilon?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, notice how that looks kinda like a limit.... Anyway...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Consider 1/2, 1/3, and 1/4 Do they have a neighborhood?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0at least its not that specific of a number

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So can you understand how these three points would be boundary points?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In this way, neighborhoods are analogous to continuity.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does my analogy sound consistent to you understanding of what a neighborhood is?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay so suppose we did have two points that were directly next to each other... they'd have to be \[ \frac{1}{n}, \frac{1}{n+1} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Notice 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And since the numerator is even and denominator is odd, we can't reduce this fraction anymore, so it can't be expressed as \[ \frac{1}{k},k\in\mathbb{N} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thus it is not in our set

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm ok that makes sense!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is sort of a proof by contradiction as to why there can't be any neighborhoods.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, so then based on the defn of int and bd, we have to first see about continuity and neighborhood

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if it is continuous then we can see that int and bd are not an empty set

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The continuity thing was more to help me understand neighborhoods than it was to help you. =)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0=D kk but it helped me learn too! lol!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think what you need to understand is that rational numbers do NOT make any neighborhoods.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you'll always be able to find some irrational number between two rational numbers.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thus 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you get something in interval notation, just remember that changing [ ] to ( ) should give you interior set.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so even union of sets

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, it should work for unions of sets well too Just not for intersections of sets.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0automatically the int is empty set

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For example: \[ [1,4]\cap [2,6] \]This gives us: \[ [2,4] \]this you can change them... \[ int([2,4]) = (2,4)\quad bd([2,4]) = \{2,4\} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So for intersections... just compute the intersection. It shouldn't be too bad.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You'd have to give me some \(\epsilon>0\) but clearly \(2+\epsilon\) is NOT in the set \(\{2\}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0kk thank you so much for helping me !!!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.