Convergence of coordinates to zero Consider a normed finite-dimensional vector space $V$ with some norm $|| .||$
Say a sequence of vectors in this vector space $v_m \rightarrow 0$ where $0$ is the zero vector.
Let $\{b_1,b_2,\ldots,b_n\}$ is a basis for $V$.
Therefore $v_m = \sum_{i=1}^{n} c_i ^m b_i$ for some scalars $c_i$
I am trying to prove that $\forall m$ the sequences $\{ c^m_i \}_{i=0}^{\infty}$
converge to the scalar $0$.
This seems like an 'obvious' theorem, but I am having difficulty proving it. 
Obviously some triangle inequality has to be used here but I don't know how.
Using both the inequalities $ | ||x||- ||y|| |\leq ||x-y|| \leq ||x|| + ||y||$
on $||\sum_{i=1}^{n} c_i ^m b_i || < \epsilon$ does not give me an upper bound
on each of the $c^i_m$ sequences. 
Any hints?
 A: You can do that by first showing that norms are equivalent in a finite dimensional space (see https://math.stackexchange.com/a/345661/10385) and then you have $M\in \Bbb R, \| \cdot\|_\infty \le M\|\cdot\|$ so if for a sequence of vectors $\left(v_n\right)_{n\in \Bbb N}$, $v_n\underset{n\to +\infty}{\longrightarrow}0$ for the norm $\|\cdot\|$, that is $\|v_n\|\underset{n\to +\infty}{\longrightarrow}0$, you also have $\|v_n\|_\infty\underset{n\to +\infty}{\longrightarrow}0$ and since $0 \le |c_i^n| \le \|v_n\|_\infty\underset{n\to +\infty}{\longrightarrow}0$, $|c_i^n|\underset{n\to +\infty}{\longrightarrow}0$ so $c_i^n\underset{n\to +\infty}{\longrightarrow}0$.

And no, it doesn't work in infinite dimensional spaces.
Take $\Bbb R\left[X\right]$ the space of polynomials.
Take $\forall P \in \Bbb R\left[X\right],\left\|P\right\|=\int\limits_0^1 \left|P(t)\right|\space dt$
And $\left(\left(1-X\right)^n\right)_{n\in \Bbb N}\in \Bbb R\left[X\right]^\Bbb N$
$\left(\left\|\left(1-X\right)^n\right\|\right)_{n\in \Bbb N}$ is decreasing and bounded below so it converges to $l$. We also have $0\le l$.
$\forall \varepsilon > 0,\left\|\left(1-X\right)^n\right\|=\int\limits_0^1 \left|\left(1-t\right)^n\right|\space dt=\int\limits_0^\varepsilon \left|\left(1-t\right)^n\right|\space dt+\int\limits_\varepsilon^1 \left|\left(1-t\right)^n\right|\space dt \le \varepsilon +\int\limits_\varepsilon^1 \left|\left(1-t\right)^n\right|\space dt\underset{n\to +\infty}{\longrightarrow}\varepsilon$
So $\forall \varepsilon >0, l \le \varepsilon$ which means $l=0$.
So $\left(1-X\right)^n\underset{n\to +\infty}{\longrightarrow}\Bbb 0$
But $\left(1-X\right)^n = \sum\limits_{k=0}^n {n \choose k} 1^{n-k}\left(-X\right)^k = 1 + \sum\limits_{k=1}^n {n \choose k} \left(-X\right)^k = 1 - X\sum\limits_{k=0}^{n-1} {n \choose k+1}\left(-X\right)^k$
So since we know that $\Bbb R\left[X\right]=\Bbb R_0\left[X\right] \oplus X\Bbb R\left[X\right]$, in the canonical basis $\left(X^n\right)_{n\in \Bbb N}$, the first coordinate will remain 1.
