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manjuthottam
could someone help me understand Topology of the Reals?!
attached is the solutions for some question relating to it. the first 5 pages deal with it
i keep getting int and bd wrong
I wish I could help, dunno enough about topology yet
ok thank you though :)
I can help reason it out if you tell me definitions
Seems like interior is just stripping away the boundery.
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.
then x is called a boundary point of S.
You understand what neighborhood is?
well for first problem a) why is the interior empty set while for b) it is the whole set with open parentheses
and yes i understand what neighborhood is
what made 1/n interior pt be special to be empty while the union of two sets interior pt is the complete whole thing
i would have guessed the ans for b) to be 3 or something
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.
Despite not understanding neighborhood rigorously, that is my reasoning for b)
And it applies to c) and d) apparently
so the answer is focusing on rational numbers?
Okay, @manjuthottam do you know what continuity is?
is it just that the limit exists?
yeah the limit exists at the function
But actually, let's talk about neighborhoods a bit ok?
ok so neighborhood is x in R. while there is an epsilon greater than zero where |x-y|< epsilon?
Yeah, notice how that looks kinda like a limit.... Anyway...
Consider 1/2, 1/3, and 1/4 Do they have a neighborhood?
at least its not that specific of a number
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.
So can you understand how these three points would be boundary points?
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?
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.
In this way, neighborhoods are analogous to continuity.
Does my analogy sound consistent to you understanding of what a neighborhood is?
Okay so suppose we did have two points that were directly next to each other... they'd have to be \[ \frac{1}{n}, \frac{1}{n+1} \]
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.
And since the numerator is even and denominator is odd, we can't reduce this fraction anymore, so it can't be expressed as \[ \frac{1}{k},k\in\mathbb{N} \]
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.
hmm ok that makes sense!
This is sort of a proof by contradiction as to why there can't be any neighborhoods.
ok, so then based on the defn of int and bd, we have to first see about continuity and neighborhood
if it is continuous then we can see that int and bd are not an empty set
The continuity thing was more to help me understand neighborhoods than it was to help you. =)
=D kk but it helped me learn too! lol!
I think what you need to understand is that rational numbers do NOT make any neighborhoods.
you'll always be able to find some irrational number between two rational numbers.
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.
If you get something in interval notation, just remember that changing [ ] to ( ) should give you interior set.
ok so even union of sets
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.
Yeah, it should work for unions of sets well too Just not for intersections of sets.
automatically the int is empty set
For example: \[ [1,4]\cap [2,6] \]This gives us: \[ [2,4] \]this you can change them... \[ int([2,4]) = (2,4)\quad bd([2,4]) = \{2,4\} \]
So for intersections... just compute the intersection. It shouldn't be too bad.
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
You'd have to give me some \(\epsilon>0\) but clearly \(2+\epsilon\) is NOT in the set \(\{2\}\)
kk thank you so much for helping me !!!!