# Point of Accumulation

1. Apr 20, 2005

I was just reading about the point of accumulation. Is it right in saying that given any sub-interval, that atleast one of these could contain ani infinite amount of points? Also why couldn't all the sub intervals contain an infinite amount of points? I came across the following: $$0.a_{1}a_{2}$$ (where a1 and a2 are sub-intervals) must be the point of accumulation. I do not understand the following statement. Is it because its a decimal?

Thanks a lot

2. Apr 20, 2005

### HallsofIvy

Staff Emeritus
"given any sub-interval, that atleast one of these could contain ani infinite amount of points" What do you mean by "at least one of them"? One of what? The definition of "point of accumulation (of a set of points) is that there exist an interval (more generally neighborhood) of the point that contains an infinite number of points in the set. Of course, I is an interval about "point of accumulation" P, any sub-interval of I, that also contains P, must also contain an infinite number of point in the set.

"I came across the following: $$0.a_{1}a_{2}$$
(where a1 and a2 are sub-intervals) must be the point of accumulation. I do not understand the following statement."
What "following statement"? You don't give any "following statement".
As far as $$0.a_{1}a_{2}$$ is concerned, I don't know what that could mean if "a1 and a2 are subintervals" rather than digits! Are you sure you copied it correctly?

3. Apr 20, 2005

### joeboo

This sounds like the proof of the Bolzano-Weierstrass Theorem

Any bounded infinite subset of $\mathbb{R}^n$ has an accumulation point.

There is nothing prohibiting all the subintervals from having an infinite number of points in common with the set. Note that the theorem doesn't suggest that the set has ONE and ONLY one accumulation point, just that there is at least one.

As for the second part of your statement, I think what is intended is the INTERSECTION of the subintervals. This is nonempty ( although this is a slightly nontrivial conclusion - it requires you to know that the reals are locally sequentially compact ). None-the-less, since the original set has an infinite number of points in common with each subinterval ( chosen in accordance with the proof of the Bolzano-Weierstrass Theorem ), and any neighborhood of the intersection will contain one of the subintervals ( as the diameter of the subintervals get smaller at the rate of $2^n$ ) - then any neighborhood of the interesection of the subintervals therefore has infinitely many points in common with the original set. In otherwords, the intersection contains an accumulation point ( and in this case, there is one and only 1 in the interesection because the diameter of the subintervals goes to 0 )

I hope that wasn't too confusing, I was trying to avoid getting into specifics. If it will help to do so, let me know.