Bounded Sequence in Euclidean Space has Convergent Subsequence/Proof 3

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:

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:

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:

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