Prove that x is an accumulation point of a set
A in a metric space X if and only if for each r > 0 the set B(x, r) \ A is infinite.
Im findinding it hard to understand what is an accumulation point, I just dont know how to start this proof.
Prove that x is an accumulation point of a set
A in a metric space X if and only if for each r > 0 the set B(x, r) \ A is infinite.
Im findinding it hard to understand what is an accumulation point, I just dont know how to start this proof.
I suppose you meant: $\displaystyle B(x,r)\cap A$ infinite.
$\displaystyle \Rightarrow)$ If $\displaystyle B(x,r_0)\cap A=\{a_1,\ldots,a_m\}$ finite, choose $\displaystyle r=\min \{d(x,a_i)\}\;(x\neq a_i)$ then, $\displaystyle (B(x,r)-\{x\})\cap A=\emptyset$ .
$\displaystyle \Leftarrow)$ Try it.
p is an accumulation point of a set, A, if and only if there exist other points of A arbitrarily close to p. For example, in the set (0, 1)U {2}, the open interval from 0 to 1 together with the single point 2, the set of accumulation points is precisely [0, 1]. 2 is NOT an accumulation point because there are no other points of the set within distance 1 of the point.
Do i do something like this..
Proof By Contradiction:
Suppose x is an accumulation point of A, and that B(x,r)∩A is finite......
I don't really know where to go from here, what is meant by accumulation point, I understand the concept... But just can't write it down mathematically...
That's a good start. Now look the distance from each point in $\displaystyle B(x,r)\cap A$ to x. Since there are only a finite number of points, none of which is equal to x, there must be a smallest non-zero distance. Now take a new B(x, r) with r smaller than that smallest distance.