Is this set (average of components below some value) closed? I guess the question is rather basic, however I would appreciate an explaination.
Given a value $c\in\mathbb{R}$, I define the set $B\subset\mathbb{R}^\infty$:
$$ B:=\{(x_1,x_2,\dots):\sum_{i\geq1}(c-x_i)\geq 0\}$$
now is the set $B$ closed in $\mathbb{R}^\infty$? (with respect to the metric induced by the maximum-norm $\|x\|_\infty$)
 A: Since $x_i\mapsto c-x_i$ is an isometry of $\Bbb R^\infty$, we can equivalently ask if $B'=\{x\in\Bbb R^\infty\mid\sum_ix_i\ge 0\}$ is closed.
It is not closed. Consider the sequence $x_i=-2^{-i}$, so that $\sum_ix_i=-1$, hence $x\notin B'$. For any $n$, we can also consider the sequence $(y_n)_i=x_i+1/n$ if $i\le n$ and $(y_n)_i=x_i$ otherwise. This is also a convergent sequence, and $\sum_i(y_n)_i=\sum_ix_i+\sum_{i=1}^n1/n=0$, but $\|x-y\|_\infty\le1/n$. Thus $(y_n)_{n\in\Bbb N}$ is a sequence of elements in $B'$ which converge to $x\notin B'$.
It is closed in the $\ell_1$ norm, however. (Note that $B$ itself will be empty for all $c\ne0$, because if $\sum_ix_i$ converges then $\sum_i(c-x_i)$ does not. But the question does make sense for $B'$, and the map $x_i\mapsto-x_i$ is an $\ell_1$ isometry, so $B$ is closed for $c=0$ iff $B'$ is closed.) If $x\notin B$, then $\sum_ix_i$ converges to some $-\epsilon<0$. But then for any $y\in\ell_1$ such that $\|x-y\|_1<\epsilon$, we have $\epsilon>\sum_i(y_i-x_i)=\sum_iy_i-\sum_ix_i=\sum_iy_i+\epsilon$, so $\sum_iy_i<0$ and hence $y\notin B'$.
