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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 $X_2^*$).2012-10-13
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    Are you asking about the statement of the theorem or its proof?2012-10-13
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    @fgp What about elements in ${X_1}^* \setminus \phi_1[X]$?2012-10-13
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    @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 $f\circ \phi_1 = \phi_2$.2012-10-13
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    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 $X$.2012-10-13
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    @Ayman, But you've left out the important point that the isometry between $X_1^*$ and $X_2^*$ is unique, so they are as "equal" as any two mathematical objects usually can be hoped to be.2012-10-13
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    @KeenanKidwell Yes indeed. The bijective isometry is unique.2012-10-13
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    @Ayman I don't understand why uniqueness of such isometry is important.2012-10-13

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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.2012-10-13
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    @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 $X_1^*\setminus X_1\ne X_2^*\setminus X_2$.2012-10-13
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    @KeenanKidwell: that is not true, as Brian M. Scott indicated. I've added an example to illustrate this.2012-10-13
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    @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 $X$.2012-10-13
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    @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.2012-10-13
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    @Keenan: I’m afraid that you’re wrong. It most definitely does not mean anything of the kind. See tomasz’s example. The codomain of the embedding $\text{id}_X$ is $X_i^*\ne X_i$; it is not $X$.2012-10-13
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    @KeenanKidwell: I meant that they are identity as functions, not as morphisms, that is, $X\subseteq X_1^*,X_2^*$. and $\varphi_{1,2}$ are the identity embeddings.2012-10-13
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    @Brian What is the codomain of $\mathrm{id}_X$? The standard usage of this notation means the identity map from $X$ to itself. If $X_1^*\neq X$, then $\phi_1$ and $\mathrm{id}_X$ don't have the same codomain, so they can't be equal. Perhaps tomasz intends $\mathrm{id}_X$ to mean the inclusion of $X$ into a larger space of which it is a subset. If so, that's fine, but I would say it is not in accordance with the way $\mathrm{id}_X$ is used in, e.g., category theory, algebra, etc..2012-10-13
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    @tomasz Thank you! I wanted to know 'letting $\phi$ designate one of the completions of $X$ where $\phi[X]$ is dense in completions of $X$' makes sense.2012-10-13
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    @tomasz I'm just accustomed to $\mathrm{id}_X$ meaning the identity arrow, which by definition has codomain $X$ (this is how it's used for an object $X$ of any category). That's why I said that, in that case, $f$ must be the identity.2012-10-13
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    @KeenanKidwell: I think it's pretty obvious what I meant from the context, and you're just being picky.2012-10-13
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    @Keenan: **Obviously** that’s what tomasz means! And this is topology, not category theory, algebra, etc. As a set-theoretic topologist I deal in functions, not morphisms, and in this context that’s obviously what tomasz is doing as well.2012-10-13
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    I don't think I'm being picky, and I disagree that it's obvious what you meant, but we can agree to disagree. I would guess that most algebraists would see the notation $\mathrm{id}_X$ and assume it meant the identity arrow from $X$ to itself. But as it has been made clear in the comments what you did intend, I'll withdraw from the conversation. I will say I don't think there's any need for the bold or accusations of pickiness. I genuinely am not accustomed to seeing the notation used in that way.2012-10-13
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    @tomasz I don't understand why uniqueness of such isometry is important since i can't use $\phi$ to designate one of completions.2012-10-13
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    @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