Hmm, I believe \(\displaystyle \frac{(-1)n}{n}=-1\).brooklysuse said:Find the set of cluster points for the set A := {(−1)n/n : n ∈ N}.
I believe 0 is a cluster point
I guess you mean $A := \{(−1)^n/n : n \in \Bbb{N}\}$. You are correct that $0$ is the only cluster point. To prove it, you will need to use the definition of a cluster point, which is ... ? (Start from there.)brooklysuse said:Find the set of cluster points for the set A := {(−1)n/n : n ∈ N}. Justify your answer with
proof.
I believe 0 is a cluster point but I can't figure out how to prove this, or how to prove any other point is not.
Any quick help would be appreciated. Thanks.
A cluster point, also known as an accumulation point, is a point in a set of numbers where an infinite number of points in the set are arbitrarily close to it. In other words, any neighborhood around the cluster point will contain an infinite number of points from the set.
To find cluster points, you must first determine the set of numbers you are working with. Then, you can use the definition of a cluster point to identify points in the set that are arbitrarily close to each other. These points will be the cluster points of the set.
While both cluster points and limit points are points in a set that are arbitrarily close to each other, there is a subtle difference between the two. A cluster point must have an infinite number of points in the set that are arbitrarily close to it, while a limit point only needs to have at least one point in the set that is arbitrarily close to it.
Yes, a set can have multiple cluster points or accumulation points. This is because there can be multiple points in a set that are arbitrarily close to each other, making them all cluster points. However, there can only be one limit point for a set.
Cluster points and accumulation points are important concepts in real analysis and topology. They are used to study the behavior of sequences and series, as well as to define the notions of continuity and compactness. They are also useful in proving theorems and solving problems in these areas of mathematics.