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Let $X, Y$ be metric spaces with distance functions denoted by $d_X, d_Y$ respectively. Consider a map $f \colon X \rightarrow Y$. I am interested in the following property: for every $x,y,z \in X$, if $d_X(x,y) \leq d_X(x,z)$ then $d_Y(f(x), f(y)) \leq d_Y(f(x), f(z))$. My question is: is this property known and studied? If yes, could you elaborate on the terminology and give references for it? How is it related to $f$ being an isometry?

Thank you in advance for your help.

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  • $\begingroup$ Note that this means preserving the 3-ary relation $R(x,y,z)$ defined by "$d(x,y)\le d(x,z)$". A metric space can be characterized by a countable number of binary relations (namely the relations $d(x,y)\le r$ for $r$ ranging over rationals), but not, as far as I know, a single relation. $\endgroup$
    – YCor
    Commented Oct 17, 2023 at 13:49
  • $\begingroup$ One can wonder which 3-ary relations $R$ arise in this way from a metric space. Obvious conditions: are $R(x,x,y)$, $R(x,y,y)$ hold for all $x,y$, transitivity of the binary relation $R(x,-,-)$ for each $x$, and $R(x,y,x)$ implies $x=y$. $\endgroup$
    – YCor
    Commented Oct 17, 2023 at 13:56
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    $\begingroup$ @PietroMajer no, it's much stronger. It indeed implies monotone, say, non-decreasing (up to compose with $x\mapsto -x$). Taking $t>0$, $y=x-t$ and $z=x+t$ and then exchanging $y$ and $z$ shows $f(x+t)-f(x)=f(x)-f(x-t)$. By continuity, this seems to force $f$ to be affine. I.e., in this case, we have only similarities (considering constant as $0$-similarities). $\endgroup$
    – YCor
    Commented Oct 18, 2023 at 4:43
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    $\begingroup$ I think this argument can be generalized to $\mathbb{R}^n$. Indeed, if $f: \ \mathbb{R}^n \rightarrow \mathbb{R}^n$ has this property, then it has to preserve circumcenters of simplices. By this I mean that if $c$ is the circumencter of a (non-degenerate) simplex with vertices $v_1, ..., v_{n+1}$, then $f(c)$ is the circumcenter of the simplex with vertices $f(v_1), ..., f(v_{n+1})$. Does this imply that $f$ is affine? $\endgroup$
    – gm01
    Commented Oct 18, 2023 at 5:00
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    $\begingroup$ Suppose $X$ is “connected by folding rules”, meaning that for any $r>0$ and $x,x’$ in $X$ there are $x=x_0,x_1,\dots, x_m=x’$ such that $d(x_i,x_{i+1})=r$ (this is the case of any normed space $X$). Then since such an $f$ maps isosceles triangles to isosceles triangles, it also verifies $d(f(x),f(y))=\phi(d(x,y))$ for some increasing function $\phi$ (and subadditive, at least for the case of $X$ a normed space). $\endgroup$
    – Pietro Majer
    Commented Oct 18, 2023 at 7:28

2 Answers 2

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" How is it related to $f$ being an isometry?"

Obviously, any isometry $f$ has the property in question.

On the other hand, not every $f$ having this property is an isometry. Indeed, let $X=Y=\{1,2,3\}$ with $d_X(x,x)=d_Y(x,x)=0$ for $x\in X$, and with $d_X(x,y)=x+y$ and $d_Y(x,y)=1$ for distinct $x$ and $y$ in $X$. Then the identity map from $X$ to $Y=X$ has the property in question but is not an isometry. $\quad\Box$

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    $\begingroup$ There are so many trivial examples of non-isometries with this property that I expect OP to be aware of them. $\endgroup$
    – YCor
    Commented Oct 17, 2023 at 17:52
  • $\begingroup$ @YCor : On the other hand, do you know what else can be said about this particular property -- especially concerning its relation with being an isometry? $\endgroup$
    – Iosif Pinelis
    Commented Oct 17, 2023 at 21:01
  • $\begingroup$ I don't really know, OP's question is quite open-ended. $\endgroup$
    – YCor
    Commented Oct 17, 2023 at 22:43
  • $\begingroup$ @YCor : Why do you think the question " How is it related to $f$ being an isometry?" is quite open-ended? What kinds of relations can there be, in principle, between two properties? The only such relations that I can see are implications: that one of the properties implies (or does not imply) the other one. If the OP knew such relations, they should have been mentioned in the OP -- but then what would be the point of question "How is it related to $f$ being an isometry?"? So, I fail to see the point of your first comment to my question. $\endgroup$
    – Iosif Pinelis
    Commented Oct 18, 2023 at 1:33
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    $\begingroup$ In this interpretation, the question is highly off-topic: examples of non-isometries already occur between two 2-point metric spaces. When I say open-ended, I mean, for instance that when some condition are given on $X$ and $Y$, the question of determining such maps might become more interesting. E.g., what are spaces $X$ for which the only $f:X\to X$ with the condition are similarities (i.e., multiply the distance by a constant)? What about Euclidean spaces? $\endgroup$
    – YCor
    Commented Oct 18, 2023 at 4:27
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Under mild conditions on the metric space $(X,d)$, we can assure that either $f$ is constant or $f$ is injective, so $f$ is either trivial or behaves similar to an isometry. Let $\simeq$ be the equivalence relation on $X$ where we set $x\simeq y$ iff $f(x)=f(y)$. Let $U$ be an equivalence class of $\simeq$, and suppose that $U$ has more than one point. Let $r_U$ be the supremum of all real numbers $r$ such that for each $x\in U$, there exists some $y\in U$ with $d(x,y)\geq r$.

Proposition: $r_A\geq\text{Diam}(A)/2$ for any set $A$ with metric $d$.

Proof: Suppose that $s<\text{Diam}(A)$. Then there are $u,v\in A$ with $d(u,v)>s$. If $x\in A$, then $d(x,u)>s/2$ or $d(x,v)>s/2$, since if $d(x,u)\leq s/2,d(x,v)\leq s/2$, then $d(u,v)\leq s$ by the triangle inequality. Q.E.D.

In particular, $r_U>0$.

If $(X,d)$ is a metric space, then we say that a subset $C\subseteq X$ is $\epsilon$-uniformly clopen if whenever $x\in C$, we have $B_\epsilon(x)\subseteq C$.

Observation: If $C\subseteq X$, then $C\subseteq X$ is $\epsilon$-uniformly clopen if and only if $C^c$ is $\epsilon$-uniformly clopen.

Proof: $C\subseteq X$ is $\epsilon$-uniformly clopen iff $\forall x,y\in X,d(x,y)<\epsilon\rightarrow(x\in C\rightarrow y\in C)$ iff $\forall x,y\in X,d(x,y)<\epsilon\rightarrow(y\in C^c\rightarrow x\in C^c)$ iff $C^c$ is $\epsilon$-uniformly clopen.

Q.E.D.

In particular, every $\epsilon$-uniformly clopen set is clopen. We say that a subset of a metric space is uniformly clopen if it is $\epsilon$-uniformly clopen for some $\epsilon>0$. We say that a metric space is uniformly connected if the only uniformly clopen sets are the empty set and the entire space.

Observation: The set $U$ is $r_U$-uniformly clopen.

Proof: Suppose that $x\in U$ and $d(x,y)\leq r_U$. Then there is some $z\in U$ with $d(x,z)\geq r_U$. Therefore, $d(f(x),f(y))\leq d(f(x),f(z))=0$ since $x\simeq z$. Therefore, $f(x)=f(y)$, so $y\in U$ as well. Q.E.D.

Therefore, if $(X,d)$ is uniformly connected and $d(x,y)\leq d(x,z)\Rightarrow d(f(x),f(y))\leq d(f(x),f(z))$, then either $f$ is injective or $f$ is constant.

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  • $\begingroup$ The terminology "uniformly connected" might be misleading because it is weaker than connected. Maybe "metrically connected"? Examples of metrically connected spaces are dense subsets of connected spaces. $\endgroup$
    – YCor
    Commented Oct 19, 2023 at 14:12
  • $\begingroup$ The term 'uniformly connected' has already been established, and it is the analogue of connectedness for uniform spaces. A uniform space $X$ is uniformly connected iff the only uniformly continuous functions $f:X\rightarrow\{0,1\}$ are the constant functions. $\endgroup$
    – Joseph Van Name
    Commented Oct 19, 2023 at 16:20
  • $\begingroup$ OK, fine. I'd like to know who coined this terminology... Wikipedia quotes Cantor 1883, I doubt it... $\endgroup$
    – YCor
    Commented Oct 19, 2023 at 16:35

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