A copy of a metric space Question:
Suppose that a metric space $(X,d)$ contains exactly four points, $X = \{ x_1 ,x_2,x_3, x_4\}$. Is it true that one can find four points $p_1,p_2,p_3,p_4 $ in some Eucledian space $ \Bbb{R^n} $ such that for any $i,j \in \{1,2,3,4\}$:
$$dist(p_i,p_j) = d(x_i,x_j)$$
Attempt:
A previous problem asked for three points in $X$ and $p_i$ in $\Bbb{R^2}$ That was straight forward but when I try to extend it to four points, there is no obvious solution. I suspect there is a method for finding one if I move to $\Bbb{R^3}$ but I'm not confidant about that. Any hints?
 A: No, not every metric space can be embedded isometrically in $\mathbb{R}^n$.
Take the four-point space $X = \{x_1, x_2, x_3, c\}$, with distance function $d(x_i, x_j) = 2$, $d(x_i, c) = 1$. This is a metric space, as could be checked axiomatically, or by viewing it as a subspace of the graph where there is an edge from each $x_i$ to $c$.
Suppose $X$ embeds isometrically into $\mathbb{R}^n$. Take the plane in which $\{x_1, x_2, x_3\}$ live: they form an equilateral triangle with side length 2 sitting in this plane. Since $c$ must be equidistant from each, it must lie somewhere on the line through the centre of the triangle, perpendicular to the plane. In particular, it does not lie on any edge of the triangle. Hence we have a nondegenerate triangle $\{x_i, x_j, c\}$ with side lengths $(1, 1, 2)$, a contradiction.
A: No.  Let x1,x2,x3 be 1 apart and x4 1/2 from the others.
Check for the triangle inequality which would fail were
1/4 used.  Perhaps if R^3 were curved an isometric
embedding would be possible.
