# Subsequences and Accumulation Values

A sequence that does not converge may still have converging subsequences.

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Concept | Content |
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Convergence | Convergent sequences have a well-defined limit. |

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Limit Inferior and Limit Superior | The largest and smallest limit of all convergent subsequences. |

Open, Closed, Compact Sets | Important notions for subsets of real numbers. |

Bolzano-Weierstrass Theorem | Every bounded sequence has at least one converging subsequence. |

Heine-Borel Theorem | The theorem connecting the concept of compactness with boundedness and closedness. |

## Study *Subsequences and Accumulation Values* #

**Definition 1** (Subsequence). *Let \((a_n)_{n\in\mathbb{N}}\) be a sequence in
\(\mathbb{K}\). Let \((n_k)_{k\in\mathbb{N}}\) be a strongly
monotonically increasing sequence with \(n_k\in\mathbb{N}\) for all \(k\in\mathbb{N}\). Then \((a_{n_k})_{k\in\mathbb{N}}\) is called
a subsequence.*

**Example 2**. Consider the sequence \((a_n)_{n\in\mathbb{N}}=(\frac1n)_{n\in\mathbb{N}}\).
Then some subsequences are given by

\((a_{n_k})_{k\in\mathbb{N}}=(a_{2k})_{k\in\mathbb{N}}=(\frac12,\frac14,\frac16,\frac18,\ldots)\);

\((a_{n_k})_{k\in\mathbb{N}}=(a_{k^2})_{k\in\mathbb{N}}=(1,\frac14,\frac19,\frac1{16},\frac1{25},\ldots)\);

\((a_{n_k})_{k\in\mathbb{N}}=(a_{2^k})_{k\in\mathbb{N}}=(\frac12,\frac14,\frac18,\frac1{16},\frac1{32},\ldots)\);

\((a_{n_k})_{k\in\mathbb{N}}=(a_{k!})_{k\in\mathbb{N}}=(1,\frac12,\frac16,\frac1{24},\frac1{120},\frac1{720},\ldots)\).

**Theorem 3** (Convergence of subsequences). *Let
\((a_n)_{n\in\mathbb{N}}\) be
a convergent sequence in \(\mathbb{K}\)
with \(\lim_{n\to\infty}a_n=a\). Then
all subsequences \((a_{n_k})_{k\in\mathbb{N}}\) of \((a_n)_{n\in\mathbb{N}}\) are also
convergent with \[\lim_{k\to\infty}a_{n_k}=a.\]*

**Attention 4**. The existence of a convergent
subsequence \((a_{n_k})_{k\in\mathbb{N}}\) does in
general __not__ imply the convergence of \((a_n)_{n\in\mathbb{N}}\). For instance,
consider \((a_n)_{n\in\mathbb{N}}=((-1)^n)_{n\in\mathbb{N}}\).
Both subsequences \[\begin{aligned}
(a_{2k})_{k\in\mathbb{N}}\,&=((-1)^{2k})_{k\in\mathbb{N}}&&=(1,1,1,1,\ldots)\\
(a_{2k+1})_{k\in\mathbb{N}}\,&=((-1)^{2k+1})_{k\in\mathbb{N}}\!\!\!\!\!\!&&=(-1,-1,-1,-1,\ldots)
\end{aligned}\] are convergent though \((a_n)_{n\in\mathbb{N}}=((-1)^n)_{n\in\mathbb{N}}\)
is divergent.

However, we can “rescue” this statement by additionally claiming that \((a_n)_{n\in\mathbb{N}}\) is monotonic.

**Theorem 5** (Subsequences of monotonic sequences).
*Let \((a_n)_{n\in\mathbb{N}}\) be
a sequence in \(\mathbb{R}\). If \((a_n)_{n\in\mathbb{N}}\) is monotonic and
there exists a convergent subsequence \((a_{n_k})_{k\in\mathbb{N}}\), then \((a_n)_{n\in\mathbb{N}}\) is convergent with
\[\lim_{n\to\infty}a_n=\lim_{k\to\infty}a_{n_k}.\]*

*Proof:* Denote \(a=\lim_{k\to\infty}a_{n_k}\). We just
consider the case where \((a_n)_{n\in\mathbb{N}}\) is monotonically
increasing (the remaining part can be done analogously to the
argumentations at the end of the proof of the Theorem about bounded and
monotonic sequences. Since \((a_{n_k})_{k\in\mathbb{N}}\) is also
monotonically increasing, we have that \(a=\sup\{a_{n_k}\,:\,k\in\mathbb{N}\}\).

Let \(\varepsilon>0\). Due to the
convergence and monotonicity of \((a_{n_k})_{k\in\mathbb{N}}\), there exists
some \(K\in\mathbb{N}\) such that for
all \(k\geq K\) holds \[a-\varepsilon<a_{n_k}\leq a.\] Now
assume that \(n\geq N=n_K\).
Monotonicity then implies that \(a-\varepsilon<a_{n_K}\leq a_n\leq a_{n_n}\leq
a\). In particular, we have that \[|a-
a_n|=a-a_n<\varepsilon.\]\(\Box\)

**Definition 6** (Accumulation value). *Let \((a_n)_{n\in\mathbb{N}}\) be a sequence in
\(\mathbb{K}\). Then \(a\in \mathbb{K}\) is called
accumulation value if there exists some subsequence \((a_{n_k})_{k\in\mathbb{N}}\) with \[a=\lim_{k\to\infty}a_{n_k}.\]*

**Attention 7** (Names). Accumulation values are often
called by other names, like *accumulation points*, *limits
points* or *cluster points*.

\(a\in \mathbb{K}\) is an accumulation value if and only if in every \(\varepsilon\)-neighbourhood of \(a\), there are infinitely many elements of the sequence \((a_{n})_{n\in\mathbb{N}}\).

**Definition 8** (Accumulation values \(\pm\infty\)). *A real sequence \((a_n)_{n\in\mathbb{N}}\) is said to have
the (improper) accumulation value \(\infty\) if it is not bounded from
above. Analogously, we define the (improper) accumulation value
\(-\infty\) if it is not bounded
from below.*

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