Calculate distance between planes in the infinite-dimensional space of convergent series Let $\mathbf{c}$ be the space of all real sequences such that the corresponding series converges, i.e.
\begin{equation}
\mathbf{c}:=\left\{(x_k)_{k\in\mathbb{N}}\ |\ \sum_{k\in\mathbb{N}}x_k\ \text{converges}\right\}
\end{equation}
I'd like to explore this space in a geometrical sense, for instance I'd like to calculate the "distance" between two "planes" or affine varieties
\begin{equation}
\pi_1: \sum_{k\in\mathbb{N}}x_k=b_1
\quad\text{and}\quad
\pi_2: \sum_{k\in\mathbb{N}}x_k=b_2
\end{equation}
or the distance between $\pi_1$ and the origin. For the latter problem, my idea was to find the $1$-dimensional subspace orthogonal to $\pi_1$, that is the subspace generated by the vector $u=(1,1,1,\dots)$, find its intersection $P$ with $\pi_1$ and then calcualte the norm of $P$ (which exists because $P\in\mathbf{c}$).
The problem is that I get $0$ as an answer (or better, sums in the calculation diverge) so I think there is a problem with applying these concepts to this space. I think the problem might also depend on the vector $u$ not belonging to $\mathbf{c}$.
My question is: how can I consistently define a notion of "distance of a plane from the origin" in the space $\mathbf{c}$ such that hyperplanes do not have zero distance from the origin or from each other?
 A: To talk about distances between planes, you need a norm on this space. A natural norm to use is 
$$
\|x\| = \sup_{n\in\mathbb{N}} \left|\sum_{k=1}^n x_k\right|
$$
This is a norm that makes the space of convergent series a Banach space. To see why, express this norm in terms of partial sums $s_n=\sum_{k=1}^n x_k$; it is just the $\ell^\infty$ norm of the sequence $(s_1, s_2, \dots)$. And since the convergence of a series means the partial sums have a limit, the space of convergent series corresponds to the closed subspace of $\ell^\infty$ that consists of convergent sequences. 
Back to your question: the distance between $\pi_1$ and $\pi_2$ is $|b_1-b_2|$. Indeed, if $s_n\to b_1$ and $s_n'\to b_2$, it follows that $\sup_n|s_n-s_n'|\ge |b_1-b_2|$. On the other hand, equality holds when both sequences are  constant, which corresponds to series 
$$
b_1+0+0+\cdots
$$
and 
$$
b_2+0+0+\cdots
$$
A: You can also notice that your space $\mathbf{c}$ is a vector subspace of $c_0 = \{(x_n)_{n\in\mathbb{N}} : \lim_{n\to\infty} x_n = 0\}$, the space of all sequences which converge to $0$.
$c_0$ usually inherits the norm from the space $\ell^\infty$:
$$\|(x_n)_{n\in\mathbb{N}}\|_\infty = \sup_{n\in\mathbb{N}}\left|x_n\right|$$
so we can also equip $\mathbf{c}$ with $\|\cdot\|_\infty$.
Notice that for $b \in \mathbb{R}$, the set  $$\pi_b = \left\{(x_n)_{n\in\mathbb{N}} \in \mathbf{c} : \sum_{n=1}^\infty x_n = b\right\}$$
has distance from the origin equal to $0$. Indeed, for any $\varepsilon > 0$, take $n_\varepsilon \in\mathbb{N}$ such that $\frac{|b|}{n_\varepsilon} < \varepsilon$ and consider the sequence
$$x = \left(\underbrace{\frac{b}{n_\varepsilon}, \ldots, \frac{b}{n_\varepsilon}}_{n_\varepsilon \text{ times}}, 0, 0, \ldots\right) \in \pi_b$$
We have $\|x\|_\infty = \frac{|b|}{n_\varepsilon} < \varepsilon$ so we conclude $d(0, \pi_b) = 0.$
Now, for any two $b_1 \neq b_2$ we have $d(\pi_{b_1}, \pi_{b_2}) = 0$. Indeed, for any $\varepsilon > 0$ there exist a sequence $x \in \pi_{b_1}$ such that $\|x\|_\infty < \frac\varepsilon2$, and a sequence $y \in \pi_{b_2}$ such that $\|y\|_\infty < \frac\varepsilon2$.
Then we have $$d(\pi_{b_1}, \pi_{b_2}) \le \|x - y\|_\infty \le \|x\|_\infty + \|y\|_\infty < \varepsilon$$ so $d(\pi_{b_1}, \pi_{b_2}) = 0$.
So is entirely plausible for two disjoint sets, such as $\pi_{b_1}$ and $\pi_{b_2}$, to have distance $0$.
