# Definition help (accumulation point/limit point)

So I'm struggling with understanding this definition (and mathematical definitions in general).

Here is how my book defines accumulation point.

DEFINITION:
Let S be a point set, and let y be a point which is not necessarily in S. We call y an accumulation point of S if in each neighborhood of y there is at least one point x which is in S and distinct from y.
-------------------------------------------------------------------------------------------
Definition for an open set from my book:

DEFINITION:
A point set S is called open if for each point x_0 of S there is some neighborhood of x_0 which belongs entirely to S.
-------------------------------------------------------------------------------------------
Definition for a neighborhood from my book:

DEFINITION:
By a neighborhood of a point x_0 we mean the set of all points x such that -h<x-x_0<h, where h is some positive number.
-------------------------------------------------------------------------------------------

I think I don't understand the definition of accumulation point because I can't understand what the definition is saying geometrically. What I mean is, say we have a candidate accumulation point, call it y. So we check in every neighborhood of y to see if there exists an x which is in the set of interest and it isn't equal to y. But isn't there an infinite amount of neighborhoods of y?

If there is an infinite amount of neighborhoods of y, how can you check each of them to see if there exists an x which is in the set of interest and it isn't equal to y?

I completely confused myself, any help would be appreciated.

What I mean is, say we have a candidate accumulation point, call it y. So we check in every neighborhood of y to see if there exists an x which is in the set of interest and it isn't equal to y. But isn't there an infinite amount of neighborhoods of y?

If there is an infinite amount of neighborhoods of y, how can you check each of them to see if there exists an x which is in the set of interest and it isn't equal to y?
Don't worry, you understand better than you think. Your're right, to show that y is an accumulation point of S we have to show that every neighborhood of y has a certain property. Clearly, we don't have time to do this one by one. Therefore we use a principal of logic called "universal generalization", which is described as follows:

If A is an arbitrary element of a set S, with no characteristics specified other than it's membership in S, and we show that A has the property P, then we have shown that all the elements of the set S have property P.

Therefore you need to start the proof with "Let (a,b) be an arbitrary interval containing the point y" and end with "and therefore, (a,b) intersects S - {y}."

Therefore you need to start the proof with "Let (a,b) be an arbitrary interval containing the point y" and end with "and therefore, (a,b) intersects S - {y}."
Let me see if I understand this. So is (a,b) the neighborhood of y? And the intersection of (a,b) and S-{y} gives us x?

And the intersection of (a,b) and S-{y} gives us x?
Yes, although the intersection may contain more than just one point, which would still be consistent with "there exists at least one point x..."

An example might help, sorry if it's too trivial, but it illustrates the proof mechanic:

Claim: The point 0 is an accumulation point of the set of real numbers.

Proof: Let (a,b) be an open interval containing 0. Note that b is not equal to 0 by definition. Therefore b/2 is greater than 0 and less than b, and so b/2 is a point not equal to zero that is contained in (a,b) , and b/2 is a real number, therefore since (a,b) is an arbitrary interval containing 0 which intersects Reals - {0} we have shown that every interval containing 0 intersects Reals - {0} and so 0 is a limit point of the set of reals.

Normally one would omit the words subsequent to "therefore since...", but they are always implicitly a part of the logic. Also notice that b/2 is just one of many points I could have chosen, b/3 , b/4, etc are all fine as well, all I had to do is exhibit a single point in (a,b) which is in Reals - {0}.

Hmm... thanks this helps a lot!

I have been having great trouble understanding limit points and came across this thread. I apologize in advance for reviving an old thread.

The discussion above is very helpful, but I feel I still don't understand the concept completely. Since I don't know how else to phrase my questions, I start with an example:

Let A=(0,1) be a subset of the set of real numbers.

Is x=1 a limit point of A? I imagine a number line with a circle centered at 1, with r>0. The "edge" of the circle sweeps across (0,1), checking at each r whether there exists a point y=/=x. The circle eventually becomes infinitesimal, and still it contains points y=/=x. So, I conclude that x=1 is a limit point of A.

But now, what if A= [0,1]? Is x=1 a limit point of A? I repeat the same process as above, and since the circle cannot have radius=0, it always contains points that aren't equal to x=1. But I have been told that in this case 1 is not a limit point. I've been trying to understand the definition of limit point but I just don't get it. Can anyone please help me figure out my misconception?

Hi Ediliter,
I think in your example of the set $S = [0,1] \subset \mathbb{R}$, we have that $1 \in S$ is still a limit point of the set $S$.
(It doesn't matter if the interval is open or closed)