# Bounded Sequence in Euclidean Space has Convergent Subsequence

## Theorem

Let $\sequence {x_i}_{i \mathop \in \N}$ be a bounded sequence in the Euclidean space $\R^n$.

Then some subsequence of $\sequence {x_i}_{i \mathop \in \N}$ converges to a limit.

## Proof 1

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

Because $\sequence {x_i}_{i \mathop \in \N}$ is bounded, we find $v \in \R^n$ and $\epsilon \in \R_{>0}$ such that:

$\forall i \in \N: \map d {v, x_i} < \epsilon$

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

Thus $\sequence {x_i}_{i \mathop \in \N}$ can be considered as a sequence in the compact metric space $\struct {S, d \restriction_{S \times S} }$.

By Compact Subspace of Metric Space is Sequentially Compact in Itself, $\sequence {x_i}_{i \mathop \in \N}$ has a convergent subsequence in $S$.

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

$\blacksquare$

## Proof 2

Let the range of $\sequence {x_i}$ be $S$.

By Closure of Bounded Subset of Metric Space is Bounded $\map \cl S$ is bounded in $\R^n$.

By Topological Closure is Closed, $\map \cl S$ 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.

$\blacksquare$

## Proof 3

We have that all norms on $\R^n$ are equivalent.

Choose the Euclidean norm $\norm {\, \cdot \,}_2$.

Proof by induction will be used.

### Basis for the induction

Let $n = 1$.

Then the proof is given by Bolzano-Weierstrass theorem.

### Induction hypothesis

Suppose, a bounded sequence $\sequence {\boldsymbol x_i}_{i \mathop \in \N}$ in $\R^n$ has a convergent subsequence $\sequence {\boldsymbol x_{i_k}}_{k \mathop \in \N}$.

Then we have to show, that the same property holds for $\R^{n + 1}$.

### Induction step

Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a bounded sequence in $\R^{n + 1}$.

Split $\mathbf x_n$ into its first $n$ components and the last one:

$\mathbf x_n := \tuple {\boldsymbol \alpha_n, \beta_n}$

where $\boldsymbol \alpha_n \in \R^n$ and $\beta_n \in \R$

Then:

 $\ds \norm {\boldsymbol \alpha_n}_2$ $\le$ $\ds \sqrt {\norm {\boldsymbol \alpha_n}_2^2 + \beta_n^2}$ $\ds$ $=$ $\ds \norm {\mathbf x_n}_2$ Definition of Euclidean Norm

Hence, $\sequence {\boldsymbol \alpha_n}_{n \mathop \in \N}$ is a bounded sequence in $\R^n$.

By induction hypothesis, it has a convergent subsequence $\sequence {\mathbf \alpha_{n_k} }_{k \mathop \in \N}$.

Denote its limit by $\alpha$.

Furthermore:

 $\ds \norm {\beta_n}_2$ $\le$ $\ds \sqrt {\norm {\boldsymbol \alpha_n}_2^2 + \beta_n^2}$ $\ds$ $=$ $\ds \norm {\mathbf x_n}_2$ Definition of Euclidean Norm

Therefore, $\sequence {\beta_n}_{n \mathop \in \N}$ is bounded.

By Bolzano-Weierstrass theorem, $\sequence {\beta_n}_{n \mathop \in \N}$ has a convergent subsequence $\sequence {\beta_{n_k}}_{k \mathop \in \N}$.

Denote its limit by $\beta$.

Then we have that:

 $\ds \lim_{k \to \infty} \mathbf x_{n_k}$ $=$ $\ds \lim_{k \to \infty} \tuple {\boldsymbol \alpha_{n_k}, \beta_{n_k} }$ $\ds$ $=$ $\ds \tuple {\boldsymbol \alpha, \beta}$ $\ds \in \R^{n + 1}$

Hence, the bounded sequence $\sequence {\mathbf x_n}_{n \mathop \in \N}$ has a convergent subsequence $\sequence {\mathbf x_{n_k}}_{k \mathop \in \N}$.

$\blacksquare$