# Bounded Sequence in Euclidean Space has Convergent Subsequence/Proof 3

## 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

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 $\sequence {\mathbf x_{n_k}}_{k \mathop \in \N}$.

$\blacksquare$