Accumulation Point

• Aug 18th 2007, 06:19 PM
ThePerfectHacker
Accumulation Point
What does it mean "accumulation point". Used in the following theorem. If two holomorphic functions in a region \$\displaystyle D\$ agree at a set of points with an accumulation point in \$\displaystyle D\$ then \$\displaystyle f=g\$ on \$\displaystyle D\$.

What does that mean? The book assumes the reader knows what "accumulation" means.
• Aug 18th 2007, 06:33 PM
shilz222
Its a concept in topology and real analysis: See here
• Aug 18th 2007, 06:40 PM
kwah
In a literary sense, accumulation means 'build-up' .. ie, an accumulation of dust had built up on ThePerfectHacker's bar of soap (sorry :p )

However, I'm unsure whether or not it has a seperate specialist mathematical meaning.

After a quick Google search (always recommended ;) ), I come across this:

Accumulation point (mathematics - encyclopedia article about Accumulation point (mathematics.)
Quote:

A prototypical example of a limit point is an accumulation point, which is a limit of a sequence.
...
Types of limit points

If every open set containing x contains infinitely many points of S then x is a specific type of limit point called a ω-accumulation point of S.

If every open set containing x contains uncountably many points of S then x is a specific type of limit point called a condensation point of S.
I apologise if this doesn't help, but I'm weary of saying more than this, but I'd rather just give you the information rather than risk giving you wrong opinions :p

Somebody else should be able to give more specific information.
• Aug 18th 2007, 06:44 PM
JakeD
Quote:

Originally Posted by ThePerfectHacker
What does it mean "accumulation point". Used in the following theorem. If two holomorphic functions in a region \$\displaystyle D\$ agree at a set of points with an accumulation point in \$\displaystyle D\$ then \$\displaystyle f=g\$ on \$\displaystyle D\$.

What does that mean? The book assumes the reader knows what "accumulation" means.

A point \$\displaystyle x\$ is an accumulation point of a set \$\displaystyle S\$ if every neighborhood \$\displaystyle N_x\$ of \$\displaystyle x\$ contains a point of \$\displaystyle S\$ other than \$\displaystyle x\$: \$\displaystyle N_x \cap S \setminus {x} \ne \emptyset.\$ For this purpose, a neighborhood of \$\displaystyle x\$ means any open set containing \$\displaystyle x\$.
• Aug 18th 2007, 06:50 PM
ThePerfectHacker
Quote:

Originally Posted by shilz222
Its a concept in topology and real analysis: See here

Of course, I looked at Wikipedia. But they use Topology. I want an elementary explanation of what it means and what the theorem says.

Quote:

Originally Posted by JakeD
A point \$\displaystyle x\$ is an accumulation point of a set \$\displaystyle S\$ if every neighborhood \$\displaystyle N_x\$ of \$\displaystyle x\$ contains a point of \$\displaystyle S\$ other than \$\displaystyle x\$: \$\displaystyle N_x \cap S \setminus {x} \ne \emptyset.\$ For this purpose, a neighborhood of \$\displaystyle x\$ means any open set containing \$\displaystyle x\$.

Okay. But "region" means an "open connected set" so does that not mean that a point is certainly an accumulation point? (Look at the theorem).
• Aug 18th 2007, 06:57 PM
shilz222
If infinitely many numbers are given in a finite interval, these numbers possess at least 1 point of accumulation. That is, at least 1 point \$\displaystyle P \$ such that in every interval about the point \$\displaystyle P \$, however small there lie infinitely many of the given numbers. Its sort of like a Dekedind cut.

Example: Subdivide the interval [0,1] into 10 equal parts (0.1 ... 0.9). One of the sub-intervals must contain infinitely many numbers. Choose this interval to be [0.1, 0.2]. Now subdivide this interval into 10 equal parts. Clearly, we find that there is an accumulation point.

So you eventually get the point of accumulation to be P = 0.a1a2a3....
• Aug 18th 2007, 07:31 PM
JakeD
Quote:

Originally Posted by ThePerfectHacker
What does it mean "accumulation point". Used in the following theorem. If two holomorphic functions in a region \$\displaystyle D\$ agree at a set of points with an accumulation point in \$\displaystyle D\$ then \$\displaystyle f=g\$ on \$\displaystyle D\$.

What does that mean? The book assumes the reader knows what "accumulation" means.

Quote:

Originally Posted by JakeD
A point \$\displaystyle x\$ is an accumulation point of a set \$\displaystyle S\$ if every neighborhood \$\displaystyle N_x\$ of \$\displaystyle x\$ contains a point of \$\displaystyle S\$ other than \$\displaystyle x\$: \$\displaystyle N_x \cap S \setminus {x} \ne \emptyset.\$ For this purpose, a neighborhood of \$\displaystyle x\$ means any open set containing \$\displaystyle x\$.

Quote:

Originally Posted by ThePerfectHacker
Okay. But "region" means an "open connected set" so does that not mean that a point is certainly an accumulation point? (Look at the theorem).

A point \$\displaystyle x\$ is an accumulation point of a set \$\displaystyle S\$ if it is not isolated in \$\displaystyle S.\$ That means you can always find a point \$\displaystyle y\ne x\$ in \$\displaystyle S\$ as close as you want to \$\displaystyle x.\$

I'm not sure about the theorem as you've stated it. Here is another version from here. Two holomorphic functions that agree at every point of an infinite set with an accumulation point inside the intersection of their domains also agree everywhere in some open set.

The theorem says essentially that two holomorphic functions that are equal on any set \$\displaystyle S\$ with a non-isolated point \$\displaystyle x\$ are equal "everywhere."
• Aug 19th 2007, 01:53 AM
Rebesques
I go with JakeD. :)

The way to practically use the theorem is to find a bounded sequence (a_n) with f(a_n)=g(a_n).

About the notion of accumulation point? It is just the set of all limits of all sequences from the original set.