whats the difference?
Can't be..
for the set A= {$\displaystyle 1/n+ 1/m$ | n and m belongs to natural numbers}
only 0 is a limit point.
0 is also accumulation point but not the only one -
$\displaystyle 1/k$, while k is natural is an accumulation point.$\displaystyle (1, 1/2, 1/3, 1/4, 1/5 etc..)$
why?
because for every enviorment of 1/k there are infinite members from the set A.
[because if you choose $\displaystyle n=k$, you get $\displaystyle 1/k + 1/m$, now as m goes larger we get closer and close to 1/k infinite amount of times]
So! what makes $\displaystyle 1/k$ definition-wise to not be a limit point??
my uni is incredibley bad at explaining this.
and indeed accumulation point is barely mentioned in books (only limit points)
so what the hell?
The only way that I can make sense of this is if you are considering A= {$\displaystyle 1/n+ 1/m$} to be not just a set but a directed net (directed by the directed set {(m,n)}). In that sense, 0 is the only limit point of A. The reason is that the definition of limit point of a directed net requires that every neighbourhood of the limit point should contain all the members of A for which both m and n are sufficiently large.
If you are considering A merely as a point set, then the points 0 and 1/k are all limit points (or accumulation points, whichever word you prefer to use), and there is nothing special to distinguish 0 from the other limit points.
accumulation point is the point where in an infinite number of points lie on its epsilon neighborhood.. whereas, the limit point is an accumulation point but only a finite points lie outside its epsilon neighborhood..
so we can say, the accumulation point doesn't really care whether there are infinite number of points outside its neighborhood, as long as there are infinite points lie in its neighborhood..
I don't understand the solution till the end.
If I take x which is between 0 and a 1. and x is not 1/integer.
And say $\displaystyle 1/(k-1)<x<1/k $ for some integer k. (doesn't it needs a prove aswell?)
then in the enviorment (1/(k-1), 1/k)
i need to show there are finite (including 0) elements from group A.
I don't understand the delta part of the solution(help?), which I believe explains it. but this is how I did this :
I say that $\displaystyle 1/n + 1/m > 1/(k-1)$ for finite numbers of n and m.
when $\displaystyle 1/n > 1/2k-2 $ AND $\displaystyle 1/m > 1/2k-2$ (***)
then indeed $\displaystyle 1/n + 1/m > 1/(k-1) $
this happens whilst
$\displaystyle n<2k-2 $ AND $\displaystyle m< 2k-2$, and this happens for finite numbers of n and m.
But I'm not sure this proves it. because:
because of the line ***. this is not the only condition for the inequality to be correct, so there might be other m and n's, perhaps infinte that are suffiecent for the inequality.
what am I missing?
I'll try to explain the argument a bit more fully.
We want to show that if 0 < x < 1 and x is not the reciprocal of an integer then x is not a limit point of the set {(1/m) + (1/n)}. Given such an x, let δ be the distance from x to the nearest number of the form 1/n. Then δ>0, and |x – (1/n)| ≥ δ for all n.
The idea is to divide the distance δ by 2 and say that if a point is within distance δ/2 of a number of the form 1/n, then it must be at a distance greater than δ/2 from x.
There are only finitely many numbers of the form (1/m) + (1/n) for which both m and n are less than or equal to 2/δ. Given any other number of the form (1/m) + (1/n), either m or n (or both) must be greater than 2/δ. Suppose that m > 2/δ. Then 1/m < δ/2. Therefore (1/m) + (1/n) is within distance δ/2 of 1/n. By the previous paragraph, the distance from (1/m) + (1/n) to x is greater than δ/2. So (1/m) + (1/n) does not lie in the neighbourhood {y:|y–x|<δ/2} (points within distance δ/2 of x).
What this says is that the neighbourhood {y:|y–x|<δ/2} contains only finitely many points of the form (1/m) + (1/n). Therefore x is not a limit point of the set {(1/m) + (1/n)}.
I hope that makes the explanation a bit clearer. It is quite a tricky argument.