Bounded Sequence in Euclidean Space has Convergent Subsequence

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Theorem

Let $\left\langle{x_i}\right\rangle_{i \in \N}$ be a bounded sequence in the Euclidean space $\R^n$.

Then some subsequence of $\left\langle{x_i}\right\rangle_{i \in \N}$ converges to a limit.


Proof 1

Denote with $d$ the Euclidean metric on $\R^n$.

Because $\left\langle{x_i}\right\rangle_{i \in \N}$ is bounded, we find $v \in \R^n$ and $\epsilon \in \R_{>0}$ such that:

$d \left({v, x_i}\right) < \epsilon$

for all $i \in \N$.


Therefore, all $x_i$ are members of the closed $\epsilon$-ball $S = B_\epsilon^- \left({v}\right)$.

By Closed Ball in Euclidean Space is Compact, $S$ is compact.

Thus $\left\langle{x_i}\right\rangle_{i \in \N}$ can be considered as a sequence in the compact metric space $\left({S, d \restriction_{S \times S}}\right)$.


By Compact Subspace of Metric Space is Sequentially Compact in Itself, $\left\langle{x_i}\right\rangle_{i \in \N}$ has a convergent subsequence in $S$.

In particular, since $S$ is a metric subspace of $\R^n$, it follows that $\left\langle{x_i}\right\rangle_{i \in \N}$ has a convergent subsequence in $\R^n$ as well.

$\blacksquare$


Proof 2

Let the range of $\left\langle{x_i}\right\rangle$ be $S$.

By Closure of Bounded Subset of Metric Space is Bounded $\operatorname{cl} \left({S}\right)$ is bounded in $\R^n$.

By Topological Closure is Closed, $\operatorname{cl} \left({S}\right)$ is closed in $\R^n$.

By the Heine-Borel Theorem, $S$ is compact.

The result follows from Compact Subspace of Metric Space is Sequentially Compact in Itself.

$\blacksquare$