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manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1attached is the solutions for some question relating to it. the first 5 pages deal with it

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1i keep getting int and bd wrong

wio
 2 years ago
Best ResponseYou've already chosen the best response.1I wish I could help, dunno enough about topology yet

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1ok thank you though :)

wio
 2 years ago
Best ResponseYou've already chosen the best response.1I can help reason it out if you tell me definitions

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Seems like interior is just stripping away the boundery.

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1Defn 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.

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1then x is called a boundary point of S.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1You understand what neighborhood is?

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1well for first problem a) why is the interior empty set while for b) it is the whole set with open parentheses

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1and yes i understand what neighborhood is

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

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1i would have guessed the ans for b) to be 3 or something

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Well, 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.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Despite not understanding neighborhood rigorously, that is my reasoning for b)

wio
 2 years ago
Best ResponseYou've already chosen the best response.1And it applies to c) and d) apparently

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1so the answer is focusing on rational numbers?

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Okay, @manjuthottam do you know what continuity is?

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1is it just that the limit exists?

wio
 2 years ago
Best ResponseYou've already chosen the best response.1yeah the limit exists at the function

wio
 2 years ago
Best ResponseYou've already chosen the best response.1But actually, let's talk about neighborhoods a bit ok?

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1ok so neighborhood is x in R. while there is an epsilon greater than zero where xy< epsilon?

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Yeah, notice how that looks kinda like a limit.... Anyway...

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

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1at least its not that specific of a number

wio
 2 years ago
Best ResponseYou've already chosen the best response.1If 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.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1So can you understand how these three points would be boundary points?

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1so 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?

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Well, 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.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1In this way, neighborhoods are analogous to continuity.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Does my analogy sound consistent to you understanding of what a neighborhood is?

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

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Notice 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.

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

wio
 2 years ago
Best ResponseYou've already chosen the best response.1So 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.

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1hmm ok that makes sense!

wio
 2 years ago
Best ResponseYou've already chosen the best response.1This is sort of a proof by contradiction as to why there can't be any neighborhoods.

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1ok, so then based on the defn of int and bd, we have to first see about continuity and neighborhood

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1if it is continuous then we can see that int and bd are not an empty set

wio
 2 years ago
Best ResponseYou've already chosen the best response.1The continuity thing was more to help me understand neighborhoods than it was to help you. =)

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1=D kk but it helped me learn too! lol!

wio
 2 years ago
Best ResponseYou've already chosen the best response.1I think what you need to understand is that rational numbers do NOT make any neighborhoods.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1you'll always be able to find some irrational number between two rational numbers.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Thus 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.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1If you get something in interval notation, just remember that changing [ ] to ( ) should give you interior set.

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1ok so even union of sets

wio
 2 years ago
Best ResponseYou've already chosen the best response.1There 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.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Yeah, it should work for unions of sets well too Just not for intersections of sets.

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1automatically the int is empty set

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

wio
 2 years ago
Best ResponseYou've already chosen the best response.1So for intersections... just compute the intersection. It shouldn't be too bad.

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1oh 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

wio
 2 years ago
Best ResponseYou've already chosen the best response.1You'd have to give me some \(\epsilon>0\) but clearly \(2+\epsilon\) is NOT in the set \(\{2\}\)

manjuthottam
 2 years ago
Best ResponseYou've already chosen the best response.1kk thank you so much for helping me !!!!
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