Prove that $a$ is a cluster point of $E$ $\iff$ for each $r>0$, $E\cap B_r(a)$ \ $\{a\}$ is nonempty.

Question

Question:

Let $E$ be a subset of $\Bbb R^n$

Prove that $a$ is a cluster point of $E$ $\iff$ for each $r>0$, $E\cap B_r(a)$ \ $\{a\}$ is nonempty.

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definiton:

A point $a \in \Bbb R^n$ is cluster point of $E$ if $E\cap B_r(a)$ contains infinitely manypoints for every $r>0$

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Please can someone prove this? Thanks.

Answer 1

For $\implies$ part:

Suppose there is some $r>0$ such that $E\cap B_r(a)$ contains no point except $a$. This leads to contradiction to your definition of cluster point.

For $\impliedby$ part:

If, $a$ is not a cluster point then there would exist a $r>0$ such that, $E\cap B_r(a)$ contains finitely many points, say $\{x_1...,x_n\}$ apart from $a$. Take, $s=\min\{d(a,x_i)\}$. Clearly $E\cap B_s(a)$ \ $\{a\}$ is empty, contradiction!

Answer 2

Let $a$ be a limit point of $E$, that is for each $r>0$ the ball $B_r(a)$ intersects $E\setminus\{a\}$. If $x_1\in B_r(a)\cap E\setminus\{a\}$, then choose $r_1< d(x_1,a)$, so $B_{r_1}(a)\cap E\setminus\{a\}$ doesn't contain $x_1$. But $B_{r_1}(a)\setminus\{a\}$ still intersects $E$, so you can choose an $x_2\in B_{r_1}(a)\cap E\setminus\{a\}$. Repeat this process and you get an infinite sequence of points in $B_{r}(a)\cap E\setminus\{a\}$.

What is actually used here is the property that the intersection of all neighborhoods of $a$ equals $\{a\}$. A topological space with this property is called a $T_1$-space.

Answer 3

Hint Just one direction needs a proof

$\Leftarrow$ by contraposition: suppose that $E\cap B_r(a)\setminus\{a\}$ contains finitely many points $x_1,\ldots,x_s$ for some $r>0$ and let $d=\min_{i}d(a,x_i)$. Prove that $E\cap B_{d/2}(a)\setminus\{a\}$ is the empty set.