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Suppose I have a metric space $A$, for which there is a notion of distance $d$. If I were to make a subset of that space, is it a given that the subset will also have its distance defined by $d$? Or do I have to show it?

edit: Essentially, If I take a subset of a metric space, do I need to say, "this subset will use the same notion of distance as the thing it was taken from"? Or is that given?

Anthony Peter
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Qwertford
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2 Answers2

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You don't have to show it and you don't have to insist that you consider the same (or more precisely: restricted) metric on the subset $B$; you might decide to consider a completely different metric on $B$ - or that you do not consider $B$ as a metric space at all! However, it is a common phrase to talk about precisely that metric, e.g. "Let $B$ be a subset of $A$, endowed with the induced metric" and this is in fact so common that people tend to drop the "endowed with the induced metric". Of course, if $x$ and $y$ are points in $A$ have a distance $d(x,y)$m this fact does not change when we learn that $x,y$ are in fact elements of $B$ even.

What you may want to show once (and only once) in your life is that if $d\colon A\times A\to\mathbb R$ is a metric on $A$ then the restriction $d|_{B\times B}\colon B\times B\to \mathbb R$ is also a metric. But that is clear without calculation from the mere structure of the axioms of metric:

  • If $d(x,y)\ge 0$ holds for all $x,y\in A$, it certainly holds for all $x,y\in B$
  • If $d(x,y)=0$ with $x,y\in A$ implies $x=y$, it certainly also implies $x=y$ if we additionally know $x,y\in B$
  • If $d(x,y)=d(y,x)$ holds for all $x,y\in A$, it certainly holds also for all $x,y\in B$
  • If $d(x,z)\le d(x,y)+d(y.z)$ holds for all $x,y,z\in A$ it certainly holds for all $x,y,z\in B$

The crux in all four points was that we only have statements of the form $\forall x\forall y\forall z(\ldots)$ and no $\exists$ is involved.

Hagen von Eitzen
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In my opinion, it is as close to given as it gets.

If $A$ is your metric space, then the metric axioms hold for all points $x,y,z\in A$. If $S\subseteq A$ is some subset of $A$, then the metric axioms (with the restricted metric) hold in particular for all $x,y,z\in S\subseteq A$.

Mankind
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