Equidistant sequence in a normed space Let $(X, \|\cdot\|)$ be a normed linear space of dimension $n < \infty$. 


*

*Is it always true that we can find a sequence of $m = n + 1$ points $x_i$ such that $\|x_i - x_j\| = c > 0$ for all $i\neq j$?

*Can we choose $m$ to be bigger than $n+1$ without violating the first property?

*In case $X$ is not finite-dimensional, does there always exist a countable sequence of $x_i$ with the desired property?
Inspired by this question.
 A: *

*In the plane we can take the vertices of an equilateral triangle. In $n=3$ the vertices of a tetrahedron. In general, the vertices of a regular $n$-simplex. The construction is described in Wikipedia like this:



Begin with a point A. Mark point B at a distance r from it, and join
  to form a line segment. Mark point C in a second, orthogonal,
  dimension at a distance r from both, and join to A and B to form an
  equilateral triangle. Mark point D in a third, orthogonal, dimension a
  distance r from all three, and join to form a regular tetrahedron. And
  so on for higher dimensions.



*It is easy to see that it is not possible in $n=1$ or $n=2$. This implies that it is not possible in higher dimensions. If it were possible in dimension $n$, delete a point and then it would be possible in dimension $n-1$.

*The quoted construction can be carried out in an infinite dimensional Hilbert space. For a general Banach space, I do not know.
Edit
After reading Ilya's comment I realized that at least in some Banach space the answer to 2. and 3. is positive. In $\mathbb{R}^n$ with the norm of the maximum there are $2^n$ different points with mutual distance equal to $1$: $(e_1,\dots,e_n)$ where $e_i=0$ or $1$. In $\ell^\infty$ this gives an uncountable set of points with mutual distance $1$. So the question is wether this is true on any Banach space. 
A: As for 3. the answer is no:


*

*P. Terenzi, Successioni regolari negli spazi di Banach, Milan J.
Math., 57 (1) (1987), 275–285.

*E. Glakousakis and S. K. Mercourakis, Examples of infinite
dimensional Banach spaces without infinite equilateral sets,
preprint, 2015, arXiv:1502.02500, to appear in Serdica Math. J.,
42 (2016).   


see also Introduction in

T. Kania, T. Kochanek, Uncountable sets of unit vectors that are separated by more than 1.

for the discussion of the problem.
A: *

*Yes. In $\mathbb{R}^n$, ${\rm conv}\ \{ x_1,\cdots, x_{n}\}$ has $n-1$ dimension so that we can find an extra point.

*If $\| \ \|$ is strict, i.e., a unit sphere is strict convex surface or a shortest path between two points is unique, then we can not find $m$ point bigger than $n+1$. Assume that $\| \ \|$ is not strict. If $x_i,\ 1\leq i\leq n+2$ satisfies the property, then convex hull of $x_i, \ 1\leq i\leq n+1$ forms $n$-dimensional tetrahedron $T$. So shortest path $[x_{n+2}x_i] $ for some $i$ must penetrates $T$. So $\|\ \|_1,\ \|\ \|_\infty$ are possible. If unit sphere of $\|\ \|$ in $\mathbb{R}^2$ is a hexagon, then it is not possible.  
