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Assume that we are in a finite dimensional vector space, with basis $\{v_1,v_2,\ldots,v_n\}$. Assume also that we have a sequnce of bounded vectors, $\{x_i\}$, that is $\|x_i\|<M$ for some real number M. We can write the sequence like this:

$\{x_i\}=\{\lambda_1^iv_1+\lambda_2^iv_2+\ldots\lambda_n^iv_n\}$. It is very plausible that for each $1\le k\le n$ we have that $\{\lambda_k^i\}$, is a bounded sequence of either real or complex numbers, that is $|\lambda_k^i|<M_k$. But I do not see how to show this. Any tips? If our space was an innerproduct space, and the vectors were orthogonal it would be easy, beacuse then we could just use pythagoras, but in a normed space, we don't have orthogonality etc., all we know is that the basis vectors are linearly independent, so the scalars are unique.

The reason I need this is that this is what I need for the bolzano-weierstrass theorem to hold for an arbitrary finite dimensional normed vector space.

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Define the norm $\lvert \cdot \rvert_1 \, \colon X \rightarrow [0, \infty)$ by \begin{equation*} \lvert x \rvert_1 =\sum_{l=1}^{n} \lvert \lambda_l \rvert, \end{equation*} where $x \in X$ is given by $x = \sum_{l=1}^{n} \lambda_l v_l$. Let us now choose a vector $x_i$ from the sequence $\{x_i \} $. Then, by definition of $\lvert \cdot \rvert_1$, \begin{equation} \lvert x_i \rvert_1 = \sum_{l=1}^{n} \lvert \lambda_{l}^{i} \rvert \geq \lvert \lambda_{k}^{i} \rvert \end{equation} for all $k \in \{ 1, \dots, n \}$.

Since $X$ is a finite dimensional vector space, both norms $\lvert \cdot \rvert_1$ and $\lvert \lvert \cdot \rvert \rvert$ are equivalent; in particular, there exists a constant $C>0$ such that \begin{equation} \lvert x \rvert_1 \leq C \lvert \lvert x \rvert \rvert \quad \textrm{ for all } x \in X. \end{equation} Combining both observations, we see that, for a fixed $k \in \{1, \dots, n \}$, \begin{equation} \lvert \lambda_{k}^{i} \rvert \leq \lvert x_i \rvert_1 \leq C \lvert \lvert x \rvert \rvert \leq C M. \end{equation} Thus, the sequence $\{ \lambda_{k}^{i}\}$ is bounded.

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