Continuity of $(a_1,\ldots,a_n) \mapsto \sum_1^n a_j e_j$ Let $e_1,\ldots,e_j$ be a basis for a finite dimensional normed vector space $X$. I wish to show that the map $(a_1,\ldots,a_n) \mapsto \sum_1^n a_j e_j$ is continuous, where $(a_1,\ldots,a_n)$ has the Euclidean metric and we consider the metric induced by the $1$-norm on $\sum_1^n a_j e_j$.
Let $x = \sum_1^n a_j e_j$ and $y = \sum_1^n b_j e_j$. I can observe the following:
$$\begin{align*}
\|x-y\|_2 &= \|\sum(a_j-b_j)e_j\| \\
 &\le \sum \|(a_j-b_j)e_j\|_2 \\
 &= \sum |a_j-b_j| \|e_j\|_2 \\
 &\le \sum |a_j-b_j| \max_{1\le j\le n} \|e_j\|_2 \\
 &= \|x-y\|_1 \max_{1\le j\le n} \|e_j\|_2.
\end{align*}$$
Now, for any $\epsilon$, I set $\delta = \frac{\epsilon}{\max_{1\le j \le n} \|e_j\|_2}$.
But I am having trouble seeing how $\|x-y\|_2 < \delta$ gets me $\|x-y\|_1 < \epsilon$ in this case.
 A: $\newcommand{\id}{\operatorname{id}}$First, $X$ is "exactly" same as $\mathbb{R}^n$ w.r.t euclidean norm.
Let $\id$ be the transformation $(a_1,\cdots,a_n)\mapsto \sum a_je_j$. What you wrote in your question, proves the continuity of $\id$. To prove continuity of $\id^{-1}$, it's sufficient to prove that the set $\left\{\frac{\|x\|_e}{\|x\|_2}\;;x\neq0\right\}$  is bounded. Suppose $S^{n-1}$ is the unit sphere of $\mathbb{R}^n$ w.r.t euclidean norm.
There are several approaches to show this, but all are common in using compactness of the unit Sphere.
One way is this :
$\id(S^{n-1})$ is  compact in $X$. Since $0\notin S^{n-1}$, therefor $0\notin \id(S^{n-1})$. It means that $0$ is the exterior point of $\id(S^{n-1})$, i.e there is $\delta>0$ such that $B_\delta(0)\cap \id(S^{n-1})=\emptyset$.
Therefore, for any $x\in X$, if $\|x\|_2<\delta$, then $\|x\|_e\neq1$. I'll prove that $\|x\|_e<1$ later (if you ask). Since $\left\{\frac{\|x\|_e}{\|x\|_2}\;;x\neq0\right\}=\left\{\|x\|_e\;;\;\|x\|_2=\frac\delta2\right\}$, so the set is bounded by $\frac2\delta$. $\qquad\qquad\square$
A: Ok, I've made a real mess of this, but I've figured it out. Simple is good.
Let $T : (a_1,\ldots, a_n) \to \sum a_j e_j$. It is obvious that $T$ is linear. We explore its continuity at zero.
Let $\|\sum a_je_j\|_1 = \sum |a_j| < \epsilon$ for any $\epsilon > 0$.
Then, $\left(\sum |a_j|\right)^2 < \epsilon^2$, and hence $\left(\sum a_j^2\right)-2\sum_{j=1}^n\sum_{k=j+1}^n |a_j a_k| < \epsilon^2$. Let
$$\delta = \sqrt{\epsilon^2+2\sum_{j=1}^n\sum_{k=j+1}^n |a_j a_k|}$$
and we have $\|(a_1,\ldots,a_n)\|_2 = \sqrt{\sum a_j^2} < \delta$ when $\sum |a_j| < \epsilon$.
