Accumulation Point

Topics: Calculus - Topology

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(definition)

Let $A\subseteq {\mathbb{R}}^n$ and $x\in {\mathbb{R}}^n$. We say that $x$ is an accumulation point of $A$ if $\forall r>0$:

$$(B_r(x)-\{x\})\cap A\ne \varnothing$$

(observation)

$x\in \text{int}(A)⟹x\text{ is an accumulation point}$

Example

Let $A=(0,1)\subseteq \mathbb{R}$.

In this case, the accumulation points are $[0,1]$ (the interior points of $A$ and its boundary points).

Proof for $0$ Let $r>0$, then we have that:

$B_r(0)-\{0\}\cap (0,1)\ne \varnothing$

…since $\frac{r}{2}\in B_r(0),\frac{r}{2}\in A,\frac{r}{2}\ne 0$.

Proof for $1$ Let $r>0$, then we have that:

$B_r(1)-\{1\}\cap (0,1)\ne \varnothing$

…since $1-\frac{r}{2}\in B_r(1),1-\frac{r}{2}\in (0,1)$ (note that $|1-\frac{r}{2}-1|=\frac{r}{2}$).

Also note that any point in $[0,1{]}^c$ (the complement set) isn’t an accumulation point.

Proof $[0,1{]}^c$ is closed, since:

$[0,1{]}^c=(\infty ,0)\cup (1,\infty )$

Then, if $x\in [0,1{]}^c⟹\exists r>0$ such that:

$B_r(x)\subseteq [0,1{]}^c$

Therefore, we have that $B_r(x)\cap (0,1)=\varnothing$.