# Why completion of a metric space $X$ is 'unique' upto isometry?

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?

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If $\phi_1 = \phi_2$, then you trivially have $X_1^* = X_2^*$, since $\phi_1$ and $\phi_2$ have the same range if they are equal ($\phi_1$ has range $X_1^*$, $\phi_2$ has range $X_2^*$). – fgp Oct 13 '12 at 21:46
Are you asking about the statement of the theorem or its proof? – Ayman Hourieh Oct 13 '12 at 21:54
@fgp What about elements in ${X_1}^* \setminus \phi_1[X]$? – Katlus Oct 13 '12 at 21:54
@Ayman I'm asking about the statement of the theorem. Even though $\phi_1 = \phi_2$, this theorem only gives information that there exists a unique bijective isometry $f:{X_1}^* \rightarrow {X_2}^*$ such that $f\circ \phi_1 = \phi_2$. – Katlus Oct 13 '12 at 21:57
What this theorem says is that if you create two completions of a metric space, you can find a bijective isometry between the two completions. Thus, the completion is unique up to an isometry. Having $\phi_1$ and $\phi_2$ be isometries means that the distance functions in $X^*_1$ and $X^*_2$ give the same values as the distance function of $X$ when applied to elements in $X$. – Ayman Hourieh Oct 13 '12 at 22:04

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.

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Dear @tomasz, In your second comment, when you write $\phi_1=\phi_2=\mathrm{id}_X$, then the uniqueness dictates that $f$ must be the identity map because the identity map $X\rightarrow X$ satisfies the conditions that $f$ must satisfy. – Keenan Kidwell Oct 13 '12 at 22:02
@Keenan: No, even if $\varphi_1=\varphi_2=\text{id}_X$, it is quite possible that $f$ is not the identity on $X_1^*\setminus X_1$, because it’s quite possible that $X_1^*\setminus X_1\ne X_2^*\setminus X_2$. – Brian M. Scott Oct 13 '12 at 22:05
@KeenanKidwell: that is not true, as Brian M. Scott indicated. I've added an example to illustrate this. – tomasz Oct 13 '12 at 22:06
@Brian, tomasz wrote $\phi_1=\phi_2=\mathrm{id}_X$, which means that $X_1^*=X=X_2^*$, because the source and target of $\mathrm{id}_X$ are both equal to $X$. – Keenan Kidwell Oct 13 '12 at 22:08
@tomasz, When you write $\mathrm{id}_X$, you mean the identity map from $X$ to itself right? The maps $\phi_i$ have source $X$ and target $X_i^*$, so if $\mathrm{id}_X=\phi_i$, then both morphisms in particular have the same domain and target. – Keenan Kidwell Oct 13 '12 at 22:09