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manjuthottamBest ResponseYou've already chosen the best response.1
attached is the solutions for some question relating to it. the first 5 pages deal with it
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
i keep getting int and bd wrong
 one year ago

wioBest ResponseYou've already chosen the best response.1
I wish I could help, dunno enough about topology yet
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
ok thank you though :)
 one year ago

wioBest ResponseYou've already chosen the best response.1
I can help reason it out if you tell me definitions
 one year ago

wioBest ResponseYou've already chosen the best response.1
Seems like interior is just stripping away the boundery.
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
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.
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
then x is called a boundary point of S.
 one year ago

wioBest ResponseYou've already chosen the best response.1
You understand what neighborhood is?
 one year ago

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

manjuthottamBest ResponseYou've already chosen the best response.1
and yes i understand what neighborhood is
 one year ago

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

manjuthottamBest ResponseYou've already chosen the best response.1
i would have guessed the ans for b) to be 3 or something
 one year ago

wioBest ResponseYou've already chosen the best response.1
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.
 one year ago

wioBest ResponseYou've already chosen the best response.1
Despite not understanding neighborhood rigorously, that is my reasoning for b)
 one year ago

wioBest ResponseYou've already chosen the best response.1
And it applies to c) and d) apparently
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
so the answer is focusing on rational numbers?
 one year ago

wioBest ResponseYou've already chosen the best response.1
Okay, @manjuthottam do you know what continuity is?
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
is it just that the limit exists?
 one year ago

wioBest ResponseYou've already chosen the best response.1
yeah the limit exists at the function
 one year ago

wioBest ResponseYou've already chosen the best response.1
But actually, let's talk about neighborhoods a bit ok?
 one year ago

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

wioBest ResponseYou've already chosen the best response.1
Yeah, notice how that looks kinda like a limit.... Anyway...
 one year ago

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

manjuthottamBest ResponseYou've already chosen the best response.1
at least its not that specific of a number
 one year ago

wioBest ResponseYou've already chosen the best response.1
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.
 one year ago

wioBest ResponseYou've already chosen the best response.1
So can you understand how these three points would be boundary points?
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
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?
 one year ago

wioBest ResponseYou've already chosen the best response.1
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.
 one year ago

wioBest ResponseYou've already chosen the best response.1
In this way, neighborhoods are analogous to continuity.
 one year ago

wioBest ResponseYou've already chosen the best response.1
Does my analogy sound consistent to you understanding of what a neighborhood is?
 one year ago

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

wioBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

wioBest ResponseYou've already chosen the best response.1
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.
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
hmm ok that makes sense!
 one year ago

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

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

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

wioBest ResponseYou've already chosen the best response.1
The continuity thing was more to help me understand neighborhoods than it was to help you. =)
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
=D kk but it helped me learn too! lol!
 one year ago

wioBest ResponseYou've already chosen the best response.1
I think what you need to understand is that rational numbers do NOT make any neighborhoods.
 one year ago

wioBest ResponseYou've already chosen the best response.1
you'll always be able to find some irrational number between two rational numbers.
 one year ago

wioBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

manjuthottamBest ResponseYou've already chosen the best response.1
ok so even union of sets
 one year ago

wioBest ResponseYou've already chosen the best response.1
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.
 one year ago

wioBest ResponseYou've already chosen the best response.1
Yeah, it should work for unions of sets well too Just not for intersections of sets.
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
automatically the int is empty set
 one year ago

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

wioBest ResponseYou've already chosen the best response.1
So for intersections... just compute the intersection. It shouldn't be too bad.
 one year ago

manjuthottamBest ResponseYou've already chosen the best response.1
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
 one year ago

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

manjuthottamBest ResponseYou've already chosen the best response.1
kk thank you so much for helping me !!!!
 one year ago
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