How would you show that a finite set has no cluster points?

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- Feb 2nd 2010, 11:25 AMfrenchguy87Finite SETS
How would you show that a finite set has no cluster points?

- Feb 2nd 2010, 12:02 PMDrexel28
Is this in a topological space? Or a metric space?

If this is a topological space, this is wrong.

Think about $\displaystyle \left(X,\mathfrak{J}\right)$ where $\displaystyle X=\left\{a,b\right\},\text{ }\mathfrak{J}=\left\{\varnothing,X\right\}$. It is easy to see that $\displaystyle b$ is a limit point of the set $\displaystyle \{a\}$. For, the only neighborhood of $\displaystyle b$ is $\displaystyle X$ which contains a point of $\displaystyle \{a\}$ distinct from itself.

In a metric space $\displaystyle X$ though, think about the set $\displaystyle E=\left\{e_1,\cdots,e_n\right\}$ and for any $\displaystyle x\in X$ take $\displaystyle \delta=\min_{1\leqslant j\leqslant n}d\left(x,e_j\right)$ and consider $\displaystyle B_{\delta}(x)$ - Feb 2nd 2010, 12:18 PMfrenchguy87
It doesn't tell you in the problem but since I need to show the statement is true, I will assume topological space.

- Feb 2nd 2010, 12:24 PMDrexel28
- Feb 2nd 2010, 12:36 PMPlato
- Feb 8th 2010, 05:50 AMfrenchguy87
- Feb 8th 2010, 07:04 AMHallsofIvy
x is a cluster point of A if every neighborhood of x contains some point of A

**other than**x. Since A contains only a finite number of points, there can be only a finite number of points of A in any neighborhood of x. That means that, in any neighborhood of x, there is a point of A, other than x itself, which is**closest**to x. Look at a neighborhood of x with radius less than that least distance.