# Proving that a metric space with 3 points can be embedded isometrically into $\mathbb{R}^2$

My definition of an isometric embedding is that if $(M_2,d_1)$ and $(M_2,d_2)$ are metric spaces, then $G:M_1 \to M_2$ is an isometric embedding if $d_2(G(x),G(y)) = d_1(x,y)$ for all $x,y \in M_1$.

I would like to show that any metric space that has 3 points can be embedded isometrically into $\mathbb{R}^2$ with the euclidean metric. My strategy has been to define the maps point by point, but it always seems that whatever first two points' map I define, I can never get the third to work. Does anyone know of a proof?

• This problem is actually in Euclid's "Elements" (constructing a planar triangle with the given three sides subject to triangle inequalities). Apr 22 '14 at 23:13
• @user123276: Why have you destroyed the question? It was good before your last edit. Apr 23 '14 at 3:02
• Please do not deface your question. As you can see, users here can (and will) just revert your edits, and furthermore, defacing your question is just plain not nice to the users who have taken the time to provide you with answers. Apr 23 '14 at 3:28
• My apologies, was not trying to destroy the question, I noticed that I had asked the wrong question and was in the process of changing it, and left a random placeholder but got busy later in the day until now. Apr 23 '14 at 4:14
• Since users have already provided answers to this question (even if it was asked unintentionally), it would be much preferred to simply ask a new question. Apr 23 '14 at 8:07

Let's write $\{x_0, x_1, x_2\}$ as your 3-point metric space. Let's say that $y_0 = d(x_0, x_1)$, $y_1 = d(x_0, x_2)$, and $y_2 = d(x_1, x_2)$. We want to define an isometric embedding of this space into $\Bbb R^2$. Because I'm lazy, let's say $f(x_0) = 0$, and $f(x_1)=(0,y_0)$. Everything works so far: $d(f(x_0),f(x_1))=y_0=d(x_0,x_1)$.
Now we want $d(f(x_0),f(x_2))=y_1$; so let's draw a circle of radius $y_1$ around $0$. If we can isometrically embed the 3-point metric space, $f(x_2)$ will have to lie on that circle. Similarly, we want $d(f(x_1),f(x_2))=y_2$; so let's draw a circle of radius $y_2$ around $(0,y_0)$. $f(x_2)$ will have to lie on this circle as well. So it suffices to find where our two circles intersect; if they intersect at all, then have $f(x_2)$ be one of the points of intersection. This is an isometric embedding.
So we need to check that the two circles intersect. But because $y_0+y_1 \geq y_2$ by the triangle inequality, this must be true (why?)
Let $X=\{A,B,C\}$ be the metric spaces with $3$ points and $a=d(B,C)$, $b=d(A,C)$ and $c=d(A,B)$.
Then pick any points $D$, $E$ and $F$ in ${\mathbb R}^2$ such that $|EF|=a$, $|DF|=b$ and $|DE|=c$ (where $|PQ|$ denotes the Euclidean distance between two points $P$ and $Q$). It exists since $a+b\leq c$, $a+c\leq b$ and $b+c\leq a$. Then the application $f$ from $X$ into ${\mathbb R}^2$ defined by $f(A)=D$, $f(B)=E$ and $f(C)=F$ is clearly an isometry.