We have discrete metric space on set $X$.
My book says without proof that if we take $X=Q$ and for second metric space $X=Z$ then we are getting isometric spaces.Can you help me figure out why.
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1Do you know a bijection between $\mathbb Q$ and $\mathbb Z?$ Is it an isometry? – Kevin Carlson Nov 18 '21 at 18:05
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What do you need to prove if you want to show that two discrete metric spaces are isometric? – quarague Nov 18 '21 at 18:06
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1In general, two discrete spaces are isometric if and only if they have the same cardinality. – Mark Saving Nov 18 '21 at 18:06
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@quarague We need to find bijective map $\phi:X_1 \to X_2$ s.t.$f_1(x_1,x_2)=f_2(\phi(x_1),\phi(x_2)$ for $x_1,x_2 \in X_1$ $f_1$is metric of $ X_1$ $f_2$ metric of $X_2$ – unit 1991 Nov 18 '21 at 18:07
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Perhaps there is a slight misunderstanding here. What exactly do you mean by a "discrete metric space"? – Lee Mosher Nov 18 '21 at 18:09
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Or, better, what does your book mean? – Lee Mosher Nov 18 '21 at 18:10
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1@LeeMosher $f(x,y)=1$ if $x \neq y$ ,$f(x,y)=0$ if $x=y.$ – unit 1991 Nov 18 '21 at 18:11
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Set $X$ with $f$ metric will be discrete metric space. – unit 1991 Nov 18 '21 at 18:11
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Okay, then go back and read the first comments again. – Lee Mosher Nov 18 '21 at 18:29
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@LeeMosher I can't remember bijection between $Q$ and $Z$. – unit 1991 Nov 18 '21 at 18:31
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For that, take a look here. – Lee Mosher Nov 18 '21 at 18:40
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@LeeMosher Yes but it is not in closed form.How I check that $f_1(x_1,x_2)=f_2(\phi(x_1),\phi(x_2)$ – unit 1991 Nov 18 '21 at 18:49
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First of all there are at least two possible meanings behind the "discrete metric space" term. I will call them weak and strong (which is not a standard nomenclature):
- A weak discrete metric space $(M,d)$ is a space where every point is open. Meaning for every point $x\in M$ there is a radius $r>0$ such that an open ball centered at $x$ of radius $r$ is precisely $\{x\}$. Or in other words: $(M,d)$ induces the discrete topology.
- A strong discrete metric space $(M,d)$ is a space such that $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise.
Obviously strong implies weak, but not vice versa (as I will show later).
Lemma. Let $(M_1,d_1)$, $(M_2,d_2)$ be two strong discrete metric spaces. Then $(M_1,d_1)$ is isometric to $(M_2,d_2)$ if and only if they are of the same cardinality.
Proof.
"$\Rightarrow$" Every isometry is a bijection, thus they have the same cardinality.
"$\Leftarrow$" Assume that $f:M_1\to M_2$ is a bijection. We will show that $f$ preserves distance and so it is an isometry. Indeed, if $x\neq y$ then $f(x)\neq f(y)$. But then $d_1(x,y)=1=d_2(f(x),f(y))$ by definition of $d_1$ and $d_2$. $\Box$
Corollary. $\mathbb{Q}$ and $\mathbb{Z}$ with strong discrete metrics are isometric.
Proof. It is well known that both $\mathbb{Q}$ and $\mathbb{Z}$ are countable (for example see this). Therefore by our lemma these are isometric, given the strong discrete metric on both side. $\Box$
Note that the lemma is not true for weak discrete metric spaces, even of the same cardinality. Consider $X=\{1,2\}$. We will define two metrics on $X$. First of all note that it is enough to define a metric on $(1,2)$ pair only. Put: $d_1(1,2)=1$ and $d_2(1,2)=\frac{1}{2}$. Then $(X,d_1)$ and $(X,d_2)$ are both weak discrete spaces, but they are not isometric. Note that $(X,d_1)$ is strong discrete but $(X,d_2)$ is not.
freakish
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Can you please elaborate more why $(X,d_1)$ and $(X,d_2)$ are weak discrete spaces at the last example? – unit 1991 Nov 18 '21 at 19:40
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@unit1991 for both spaces pick $r=\frac{1}{3}$. Then the open ball around any point of radius $r$ is precisely the single point. – freakish Nov 19 '21 at 07:23