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

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- Feb 2nd 2010, 12:25 PMfrenchguy87Finite SETS
How would you show that a finite set has no cluster points?

- Feb 2nd 2010, 01:02 PMDrexel28
Is this in a topological space? Or a metric space?

If this is a topological space, this is wrong.

Think about where . It is easy to see that is a limit point of the set . For, the only neighborhood of is which contains a point of distinct from itself.

In a metric space though, think about the set and for any take and consider - Feb 2nd 2010, 01:18 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, 01:24 PMDrexel28
- Feb 2nd 2010, 01:36 PMPlato
- Feb 8th 2010, 06:50 AMfrenchguy87
- Feb 8th 2010, 08:04 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.