Finite SETS

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Feb 2nd 2010, 11:25 AM
frenchguy87

Finite SETS

How would you show that a finite set has no cluster points?
Feb 2nd 2010, 12:02 PM
Drexel28

Quote:

Originally Posted by frenchguy87

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

Is this in a topological space? Or a metric space?

If this is a topological space, this is wrong.

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

In a metric space $X$ though, think about the set $E=\left\{e_1,\cdots,e_n\right\}$ and for any $x\in X$ take $\delta=\min_{1\leqslant j\leqslant n}d\left(x,e_j\right)$ and consider $B_{\delta}(x)$
Feb 2nd 2010, 12:18 PM
frenchguy87

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 PM
Drexel28

Quote:

Originally Posted by frenchguy87

It doesn't tell you in the problem but since I need to show the statement is true, I will assume metric space.

(Wondering)
Feb 2nd 2010, 12:36 PM
Plato

Quote:

Originally Posted by frenchguy87

It doesn't tell you in the problem but since I need to show the statement is true, I will assume topological space.

In this case you need to tell us the general setting for the question.
What course is this for, its name?
What is the definition of cluster point?
Feb 8th 2010, 05:50 AM
frenchguy87

Quote:

Originally Posted by Drexel28

In a metric space $X$ though, think about the set $E=\left\{e_1,\cdots,e_n\right\}$ and for any $x\in X$ take $\delta=\min_{1\leqslant j\leqslant n}d\left(x,e_j\right)$ and consider $B_{\delta}(x)$

I'm not sure where to go from there...
Feb 8th 2010, 07:04 AM
HallsofIvy

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.