# Accumulation points and intervalsHow 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...

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]

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

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

so

b-a < kN

and

(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)

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?