# metric space

• Jan 27th 2008, 09:15 AM
patricia-donnelly
metric space
Hello
• Jan 27th 2008, 01:16 PM
ThePerfectHacker
Quote:

Originally Posted by patricia-donnelly
Let (X,d) be a metric space and say A is a subset of X. If x is an accumulation point of A, prove that every r-neighbourhood of x actually contains an infinite number of distinct points of A (where r>0).

Suppose $\displaystyle x$ is an accumulation point of a set $\displaystyle A$. Then there exists distinct $\displaystyle x_n \in A$ such that $\displaystyle x_n \to x$. Let $\displaystyle r>0$ then by definition of convergence it means $\displaystyle d(x,x_n) < r$ for $\displaystyle n\geq N$. Thus, $\displaystyle x_N,x_{N+1},...$ are distinct elements which lie in $\displaystyle B(x,r)$ and hence there are infinitely many.

Quote:

Using this, prove that any finite subset of X is closed.
A set is closed if and only if it contains all its accumulation points. A finite set clearly has no accumulation points, hence the set is empty. And it contains the empty set.
• Jan 28th 2008, 08:34 AM
patricia-donnelly
You've been a great help. Thank you very much