= Solution
We first justify the maximum in the proposed <supremum norm>. If a <null sequence> $x$ is nonzero, choose an index $j$ with $|x_j|>0$. All sufficiently late terms have absolute value smaller than $|x_j|/2$, since $x_i\to0$. Hence the largest absolute value occurs among a finite initial set of terms, and is finite. For the zero sequence the maximum is zero.
The <norm> axioms now follow directly. The maximum is nonnegative and vanishes precisely when every coordinate vanishes. For a scalar $a$, $\|ax\|_\infty=|a|\|x\|_\infty$. For two <null sequences>, coordinatewise addition remains a <null sequence> and
$$
|x_i+y_i|\leq|x_i|+|y_i|\leq\|x\|_\infty+\|y\|_\infty.
$$
Taking the maximum proves the <triangle inequality>, so this really is a <norm> on $V$.
Let $e_n$ have a single $1$ in position $n$. If $m\ne n$, then $\|e_m-e_n\|_\infty=1$. A convergent sequence in a <normed vector space> must be a <Cauchy sequence>, by the <triangle inequality>. These distances violate the Cauchy condition with, for example, $\varepsilon=1/2$. Thus \b[$(e_n)$ does not converge in this norm], even though each fixed coordinate converges to zero.
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