anonymous
  • anonymous
could someone help me understand Topology of the Reals?!
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
attached is the solutions for some question relating to it. the first 5 pages deal with it
1 Attachment
anonymous
  • anonymous
i keep getting int and bd wrong
anonymous
  • anonymous
I wish I could help, dunno enough about topology yet

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anonymous
  • anonymous
ok thank you though :)
anonymous
  • anonymous
I can help reason it out if you tell me definitions
anonymous
  • anonymous
Seems like interior is just stripping away the boundery.
anonymous
  • anonymous
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.
anonymous
  • anonymous
then x is called a boundary point of S.
anonymous
  • anonymous
You understand what neighborhood is?
anonymous
  • anonymous
well for first problem a) why is the interior empty set while for b) it is the whole set with open parentheses
anonymous
  • anonymous
and yes i understand what neighborhood is
anonymous
  • anonymous
what made 1/n interior pt be special to be empty while the union of two sets interior pt is the complete whole thing
anonymous
  • anonymous
i would have guessed the ans for b) to be 3 or something
anonymous
  • anonymous
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.
anonymous
  • anonymous
Despite not understanding neighborhood rigorously, that is my reasoning for b)
anonymous
  • anonymous
I mean for a)
anonymous
  • anonymous
And it applies to c) and d) apparently
anonymous
  • anonymous
@manjuthottam make sense?
anonymous
  • anonymous
so the answer is focusing on rational numbers?
anonymous
  • anonymous
Okay, @manjuthottam do you know what continuity is?
anonymous
  • anonymous
is it just that the limit exists?
anonymous
  • anonymous
yeah the limit exists at the function
anonymous
  • anonymous
But actually, let's talk about neighborhoods a bit ok?
anonymous
  • anonymous
ok so neighborhood is x in R. while there is an epsilon greater than zero where |x-y|< epsilon?
anonymous
  • anonymous
Yeah, notice how that looks kinda like a limit.... Anyway...
anonymous
  • anonymous
Consider 1/2, 1/3, and 1/4 Do they have a neighborhood?
anonymous
  • anonymous
i'm guessing no?
anonymous
  • anonymous
at least its not that specific of a number
anonymous
  • anonymous
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.
anonymous
  • anonymous
So can you understand how these three points would be boundary points?
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
In this way, neighborhoods are analogous to continuity.
anonymous
  • anonymous
hmm ok
anonymous
  • anonymous
Does my analogy sound consistent to you understanding of what a neighborhood is?
anonymous
  • anonymous
yes it does
anonymous
  • anonymous
Okay so suppose we did have two points that were directly next to each other... they'd have to be \[ \frac{1}{n}, \frac{1}{n+1} \]
anonymous
  • anonymous
yes i see
anonymous
  • anonymous
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.
anonymous
  • anonymous
And since the numerator is even and denominator is odd, we can't reduce this fraction anymore, so it can't be expressed as \[ \frac{1}{k},k\in\mathbb{N} \]
anonymous
  • anonymous
thus it is not in our set
anonymous
  • anonymous
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.
anonymous
  • anonymous
hmm ok that makes sense!
anonymous
  • anonymous
This is sort of a proof by contradiction as to why there can't be any neighborhoods.
anonymous
  • anonymous
ok, so then based on the defn of int and bd, we have to first see about continuity and neighborhood
anonymous
  • anonymous
if it is continuous then we can see that int and bd are not an empty set
anonymous
  • anonymous
The continuity thing was more to help me understand neighborhoods than it was to help you. =)
anonymous
  • anonymous
=D kk but it helped me learn too! lol!
anonymous
  • anonymous
I think what you need to understand is that rational numbers do NOT make any neighborhoods.
anonymous
  • anonymous
you'll always be able to find some irrational number between two rational numbers.
anonymous
  • anonymous
ok
anonymous
  • anonymous
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.
anonymous
  • anonymous
If you get something in interval notation, just remember that changing [ ] to ( ) should give you interior set.
anonymous
  • anonymous
ok so even union of sets
anonymous
  • anonymous
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.
anonymous
  • anonymous
Yeah, it should work for unions of sets well too Just not for intersections of sets.
anonymous
  • anonymous
so like e)
anonymous
  • anonymous
automatically the int is empty set
anonymous
  • anonymous
For example: \[ [1,4]\cap [2,6] \]This gives us: \[ [2,4] \]this you can change them... \[ int([2,4]) = (2,4)\quad bd([2,4]) = \{2,4\} \]
anonymous
  • anonymous
So for intersections... just compute the intersection. It shouldn't be too bad.
anonymous
  • anonymous
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
anonymous
  • anonymous
yeah
anonymous
  • anonymous
You'd have to give me some \(\epsilon>0\) but clearly \(2+\epsilon\) is NOT in the set \(\{2\}\)
anonymous
  • anonymous
kk thank you so much for helping me !!!!

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