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

$\blacksquare$

## Sources

- 2017: Amol Sasane:
*A Friendly Approach to Functional Analysis*... (previous) ... (next): Chapter $\S 1.5$: Normed and Banach spaces. Compact sets