4
$\begingroup$

Let $(X,d)$ be a metric space. Let $({X_1}^*, {d_1}^*)$ and $({X_2}^*, {d_2}^*)$ be completions of $(X,d)$ such that $\phi_1:X\rightarrow {X_1}^*$ and $\phi_2:X\rightarrow {X_2}^*$ are isometries. ($\phi_1[X]$ and $\phi_2[X]$ are dense in ${X_1}^*$ and ${X_2}^*$ respectively)

Then, there exists a unique bijective isometry $f:{X_1}^* \rightarrow {X_2}^*$ such that $f\circ \phi_1 = \phi_2$.

Here, let $\phi_1=\phi_2$. It doesn't seem to me that ${X_1}^* = {X_2}^*$.

What is 'unique' this theorem referring to?

  • 0
    @Ayman I don't understand why uniqueness o$f$ s$u$ch isometry is important.2012-10-13

1 Answers 1

4

It means exactly what you've written: that $X_1^*$ and $X_2^*$ are actually pretty much the same thing, including the way $X$ embeds in them.

It doesn't mean that $X_1^*=X_2^*$. Even if $\phi_1=\phi_2$ (and even if $\phi_1=\phi_2=\operatorname{id} _X$ are identity), there is no reason for $f$ to be identity map.

Indeed if we take $X=[0,1)$ with Euclidean metric then you can choose some arbitrary $x_0\notin [0,1]$ and put $X_1^*=[0,1]$, $X_2^*=X\cup\{x_0\}$ with $d_1^*$ and $d_2^*$ the obvious metrics, with $\varphi_1=\varphi_2=\operatorname{id}_X$. Then $f(1)=x_0$, $f(x)=x$ elsewhere is the unique isometry, but not identity.

Furthermore, even if $X_1^*=X_2^*$ as a set, $f$ need not be identity. For example, consider a minor refinement of the above example with $X=(0,1),X_1^*=X_2^*=[0,1]$, with $X,X_1^*$ with Euclidean metric, and $X_2^*$ with almost Euclidean metric, except that it sees $1$ as $0$ and vice versa.

  • 0
    @Katlus: it's important for a similar reason to uniqueness of other kinds of completions: its purpose is so that you can talk about "the" completion of a metric space, without worrying about *which* one it is, since they're all isomorphic anyway.2012-10-13