Accumulation points and intervals
How can be proved that the limit or accumulation points of a finite interval are, beside the points inside the interval, its endpoints, no matter if the interval is closed or open.
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I edited your post to only contain one question:
Consider the interval from a to b.
Our interval might be closed or open or neither, so the possibilities are:
[a,b], [a,b), (a,b], (a,b)
We want to show that no matter which of these is our interval, the accumulation points are [a,b] (nothing more, nothing less)
There are three things we need to show:
1) every point in (a,b) is an accumulation point
2) the points a, b are accumulation points
3) every point NOT in [a,b] is NOT an accumulation point
1) Pick some number c in (a,b). Then b-c>0. Also, `c+(b-c)` might not be in the interval, because our interval might not contain b. But `c+(b-c) * "(fraction)"` will be in our interval, if we multiply by a positive fraction that is even a little bit smaller than 1. So Consider the sequence:` `
`c+ (b-c)/2, " " c+(b-c)/4, " " c+(b-c)/8, " " c+(b-c)/(16), ... `
All of these points are in the interval, and the points are getting closer and closer to c. So c is an accumulation point.
2) We can play the same game with our endpoints. Consider the sequence:
`a+(b-a)/2, " " a+ (b-a)/4, " " a+(b-a)/8, " " a+ (b-a)/(16), ...`
`a` might not be in our interval, but as soon as you add even just a little bit to `a` you are in the interval. You can't add `(b-a)` , though, because `b` also might not be in the interval. Thus, you want to add numbers greater than 0 and smaller than `(b-a)` in order to be sure you are in the interval.
All of the numbers `a+(b-a)/n` are in the interval, and they have `a` as an accumulation point. So `a` is an accumulation point of the interval.
Similarly, consider the sequence:
`b-(b-a)/2, b-(b-a)/4, b-(b-a)/8, b- (b-a)/(16), ...`
(sorry about the no spaces... when I add in spaces they oddly turn into + signs)
These points are all in the interval, and they have `b` as an accumulation point, so b is an accumulation point of the interval.
3) Finally, let c be something NOT in the interval. Then either c<a or c>b. If c<a, then every point in the interval is at least (a-c) away from c. The points don't get arbitrarily close to c; all of them are a distance of at least (a-c) away. Similarly, if b<c, then all of the points are at least (c-b) away from c. Again, they don't get arbitrarily close to c, they stay a distance of at least (c-b) away. So c is NOT an accumulation point.
So, for the (open or closed or neither) finite interval from a to b, the accumulation points are [a,b]
PS. I should have mentioned this. I was using a sequence definition. To transform the sequences into something that fits your definition, would look like the following...
Suppose we were trying to prove that a is an accumulation point of (a,b) using the definition:
a is an acc point of A iff for every k>0 there is some x in A such that |a-x|<k
Let k>0, arbitrary
we want to find x in (a,b) such that |a-x|<k
Consider (b-a)/k (this is a positive number, since b>a and k>0 )
Find some integer N that is larger:
(b-a)/k < N
b-a < kN
(b-a)/N < k
Let x = a+(b-a)/N
Then |x-a| = |(b-a)/N| < k , and x is in (a,b)
Since k was arbitary, we can do this for any k>0
So, by our definition of accumulation point, a is an accumulation point of (a,b)
If your interval is (a,oo), the accumulation points are [a,oo)
If your interval is [a,oo), the accumulation points are [a,oo)
If your interval is (-oo,b), the accumulation points are (-oo,b]
If your interval is (-oo,b] , the accumulation points are (-oo,b]
If your interval is (-oo,oo), the accumulation points are (-oo,oo)
For each of these proofs, you can use exactly the methods as above:
For example, take (a,oo)
You want to show 3 things:
1) for every point c in (a,oo), c is an accumulation point
2) a is an accumulation point
3) for every point c in (-oo,a), c is not an accumulation point
You can show these things just as in the first part:
1) c>a, so consider the sequence
`c - (c-a)/2, c-(c-a)/4, c-(c-a)/8, ...`
2) Consider the sequence
`a+1/2, a+1/4, a+1/8,...`
3) pick c in (-oo,a). Then a>c and no points in the interval will ever be closer to c than a-c
Thanks a lot for your further details. I think I understood the demo.
Good point!!! Actually, my problem were the endpoints. As the definition of the accumulation point says:
x acc point of A if and only if
for every k>0 there is a € A: |x - a| < k
As |x - a| gives the idea of "around" x and there are no points before or after the endpoints, the definitions was not totally accomplished.
Fortunately, your demo goes further of the definition and I think it's valid to proof what I was trying to proof. Thanks a lot for that!!!
In case of an interval having an infinitive lenght, I guess the same cannot be applied, at least for one of the endpoint which, necessarily, has to be infinite, right?
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