# Spooky geometry

Every separable metric space is isometrically embeddable in $\ell^\infty(\Bbb{N})$.

## 2 thoughts on “Spooky geometry”

1. Just an idea of why this should hold:
Suppose you have a *countable* metric space. Then its points are a sequence $\{x_n\}$. Take the first point and put it anywhere in $\mathbb{R}$. Take the second point and put it also in $\mathbb{R}$ in such a way that it respects the distance from the first point. If we want to do this with the third point it will probably be the case that there is not enought “room” in $\mathbb{R}$. But if we add a dimension we see that we can put the first three points in $\mathbb{R}^2$, respecting the distances. Then keep on adding a dimension each step.

As a matter of fact doing the above construction for a dense countable set we can actually construct the isometric embedding in the general case (notice that the function defined on the dense countable set is uniformly continuous and that $l^{\infty}(\mathbb{R})$ is a complete metric space.)

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2. Let $X$ be a separable metric space and let $\{x_n\}$ be a countable, dense subset. Define $f:X\to \ell^\infty(\Bbb{N})$ as $f(x) = (d(x,x_n) - d(x_n,x_1))_{n\in\Bbb{N}}$. This map is well defined since $|d(x,x_n)-d(x_n,x_1)| \leq d(x,x_1)$. We will now prove it is an isometry.

Let $x,y\in X$ and consider the $n$-th coordinates of $f(x)$ and $f(y)$. The absolute value of their difference is $|d(x,x_n) - d(y,x_n)| < d(x,y)$, and so $d_\infty(f(x),f(y))\leq d(x,y)$. On the other hand, since $x_n$ is dense, there is a sequence $x_{n_j}\longrightarrow y$, and so $|d(x,x_{n_j})-d(y,x_{n_j})|\longrightarrow d(x,y)$. This proves $d_\infty(f(x),f(y))\geq d(x,y)$ and so $f$ is an isometry.

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