accumulation point

Accumulation point

We call $a\in \mathbb{C}$ an accumulation point of $(a_{n_k}{)}_{k\in \mathbb{N}}$ if there exists a subsequence $(a_{n_k}{)}_{k\in \mathbb{N}}$ of $(a_n{)}_{n\in \mathbb{N}}$ with

$$\underset{k\to \infty }{\lim }{a}_{n_k}=a$$

Equivalently, we may use the definitions

$$\forall ϵ>0\forall n_0\in \mathbb{N}\exists n\ge n_0:|a-a_n|<ϵ,$$

or

$$\forall ϵ>0\forall n_0\in \mathbb{N}\exists n\ge n_0:a_n\in U_{ϵ}(a),$$

i.e., there are infinitely many terms of $(a_n)$ in every epsilon neighbourhood of $a$.

Compare with convergence: convergence means the sequence eventually stays within any $\epsilon$-ball, however small. An accumulation point means the sequence keeps coming back, however small the $\epsilon$-ball. Convergence is a special case: if $a_n\to a$, then $a$ is the only accumulation point.

$a_n=(-1{)}^n$ has accumulation points $\{-1,1\}$

The sequence just alternates between $-1$ and $1$. Both values appear infinitely often, so both are accumulation points. No other value is: e.g. the $\epsilon$-ball $U_{0.5}(0)$ contains zero terms.

$b_n=(-1{)}^n\left(1-\frac{1}{n}\right)$ has two accumulation points

$b_{2k}=1-\frac{1}{2k}\to 1$ and $b_{2k-1}=-(1-\frac{1}{2k-1})\to -1$

So $1$ and $-1$ are both accumulation points. Every $\epsilon$-ball around $1$ contains infinitely many terms (all even-indexed ones eventually), and same for $-1$.

Since there are two accumulation points, $(b_n)$ is divergent.

Convergence criterion

$$\underset{n\to \infty }{\lim }{a}_n\text{ exists}⟺\underset{n\to \infty }{\mathrm{liminf}}{a}_n=\underset{n\to \infty }{\mathrm{limsup}}{a}_n=\underset{n\to \infty }{\lim }{a}_n$$

Equivalently: a bounded sequence converges iff it has exactly one accumulation point.

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A sequence can have uncountably many accumulation points

Let $(a_n)$ be any enumeration of the rational numbers in $[0,1]$ (this is possible since $\mathbb{Q}$ is countable). Since the rationals are dense in $\mathbb{R}$, every $x\in [0,1]$ has rationals arbitrarily close to it, so every $x\in [0,1]$ is an accumulation point.

The set of accumulation points $[0,1]$ is uncountable and therefore contains more elements than the sequence itself.

→ Accumulation points don’t have to be in the sequence.