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my question is the following: I assume a finite metric space $A:=\{a_0,..,a_n\}$ with distances in the reals.

Assume an arbitrary $\epsilon>0$. Is it possible to find some 1-point metric extension of $A$, say $B:=A\cup\{b_0\}$ such that the following holds:

$d(b_0,a_0)<\epsilon$ and $d(b_0,a_i)\in\mathbb{Q}$ for all $i=1,..,n$?

We are not working within some big given metric space, so any abstract metric space $B$ with the described properties will work for me. I lost track between all the triangle inequalities...

Thanks a lot in advance for considering the question.

Sofie
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    Your second sentence is ambiguously worded imo. Do you mean, "For any given epsilon>0 the following holds: ...", or do you mean, "For any given epsilon>0 the following holds: ..." – Adam Rubinson Aug 18 '13 at 16:12
  • Also, take your metric space to be (R,|.|). if a_0 = sqrt(2) and a_1 = 1 then obviously there is no real number b_0 such that d(a_0,b_0) and d(a_1,b_1) are both rational. – Adam Rubinson Aug 18 '13 at 16:15
  • Hello Adam, thanks for your comments. I edited the question to avoid misunderstandings concerning epsilon. As for your example, the distance between $a_0$ and $b_0$ can be irrational. Furthermore, there is no need to find b_0 within the reals. It can be an abstract elements I only want triangle inequalities to hold. – Sofie Aug 18 '13 at 16:27
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    @AdamRubinson: your observation doesn't invalidate the question: the new metric space $B$ is not required to be a subspace of some given metric space. The required $b_0$ can be found in $\mathbb{R}^2$ with the usual metric in your example. – Rob Arthan Aug 18 '13 at 16:28
  • I see, Rob. Based on the change/clarification to the question, d(b_0,a_0) must equal 0. Since B is a metric space, we must have b_0 = a_0. – Adam Rubinson Aug 18 '13 at 16:40
  • For those familiar with Katetov-maps, the question can be rephrased like that: For a given finite metric space $A$ and $f\in E(A)$ (= the Katetov-space over $A$), does there exist some $g\in E(A)$ such that $g(A)\subset \mathbb{Q}$ and $d(f,g)\leq \epsilon$? – Sofie Aug 18 '13 at 16:43
  • @AdamRubinson: Why does $d(b_0,a_0)$ have to be zero? We assume an arbitrary $\epsilon>0$ and ask for the existence of a certain one-point metric extension depending on the already fixed $\epsilon$. We do not ask for some one-point metric extension such that the conditions hold for any $\epsilon$. – Sofie Aug 18 '13 at 16:53

2 Answers2

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Yes, you can do this. Let $m=\min_{i\ge 1} (a_i,a_0)$ and $M=\max_{i\ge 1} (a_i,a_0)$. Choose $d(b_0,a_0)<\min(\epsilon,m)$. You'll need a function $f:[m,M]\to [0,d(b_0,a_0)]$ such that:

  1. $f$ is decreasing
  2. $x\mapsto x+f(x)$ is increasing
  3. $x+f(x)\in \mathbb Q$ when $x=d(a_i,a_0)$ for some $i=1,\dots,n$.

To construct such a function, begin with $x\mapsto d(b_0,a_0)-c(x-m)$ where $c>0$ is small, and slightly perturb it at the points $x=d(a_i,a_0)$, keeping it piecewise linear in between.

For $i\ge 1$, define $$d(b_0,a_i) = d(a_0,a_i)+f(d(a_0,a_i)) $$ The rationality is built in (property 3). Let's check the triangle inequalities involving $b_0$. Everywhere below $i,j\ge 1$.

  1. $d(b_0,a_0)\le d(b_0,a_i)+d(a_0,a_i)$ holds because $d(b_0,a_0)<m\le d(a_0,a_i)$.
  2. $d(a_0,a_i)\le d(a_0,b_0)+d(a_i,b_0)$ holds because $f\ge 0$
  3. $d(b_0,a_i)\le d(b_0,a_0)+d(a_i,a_0)$ holds because $f\le d(b_0,a_0)$.

  4. $d(b_0,a_i)\le d(b_0,a_j)+d(a_i,a_j)$ is separated into two cases:

    • when $d(a_0,a_i)\le d(a_0,a_j)$, this holds by 2, even in the stronger form $d(b_0,a_i)\le d(b_0,a_j)$
    • when $d(a_0,a_i)\ge d(a_0,a_j)$, this holds by 1: add the inequalities $d(a_0,a_i)\le d(a_0,a_j)+d(a_i,a_j)$ and $f(d(a_0,a_i))\le f(d(a_0,a_j))$.
user90090
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By rescaling if necessary, make all non-zero distances in $A$ greater than $100$, and make absolute values of differences of distinct distances greater than $100$. This is not particularly important, but helps in the visualization.

Choose a positive integer $M$ such that $\frac{1}{M}\lt \epsilon$, and declare the distance of $b_0$ from $a_0$ to be $\frac{1}{M}$.

Line up $a_1,a_2, \dots$ in non-increasing order of their distances from $a_0$. Note that there may be ties.

Consider first simultaneously all the $a_i$ which are at maximum distance from $a_0$. If $a_i$ is such a point, declare the distance of $b_0$ from $a_i$ to be some rational number $r$ in the interval $d(a_0,a_i)+\frac{1}{(n+1)M}\lt r\lt d(a_0,a_i)+\frac{1}{nM}$.

Continue in this way, the next time using the adjustment $\frac{1}{nM}$ to $\frac{1}{(n-1)M}$, and so on.

The triangle inequality: There are three types of triangle that have $b_0$ as a vertex. The triangle could have (i) vertices $b_0,a_0,a_i$ or (ii) vertices $b_0,a_i,a_j$ where $d(a_0,a_i)=d(a_0,a_j)$ or (iii) vertices $(b_0, a_i,a_j)$ where $d(a_0,a_i)\ne d(a_0,a_j)$.

Type (i): By the choice of distances, $b_0a_i$ is the long side, and $d(b_0,a_i)\lt d(a_0,b_0)+d(a_0,a_i)$.

Type (ii): Long side $a_ia_j$ is no problem, since this side has length $\le d(a_0,a_i)+d(a_0,a_j)$, and we have $d(b_0,a_j)\gt d(a_0,a_j)$.

And long side $b_0a_i$ is definitely not a problem, since $d(b_0,a_j)=d(b_0,a_i)$.

Type (iii): Again, long side equal to $a_ia_j$ is not a problem, from the triangle inequality in $A$.

Suppose the long side is $b_0a_i$. By our initial scaling, this long side is much longer than $b_0a_j$. By the extension of distance to $b_0$, we have $d(b_0,a_i)-d(a_0,a_i)\lt d(b_0,a_j)-d(a_0,a_j)$. It follows that $$d(b_0,a_i)\lt d(b_0,a_j)+d(a_0,a_i)-d(a_0,a_j)\le d(b_0,a_j)+d(a_i,a_j).$$

André Nicolas
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  • @user90090 and AndréNicolas: Thanks a lot both of you for your correct and nice answers. It helped to prove that Urysohns space is separable. And I can stop manipulating triangle inequalities. Thank you once more! – Sofie Aug 18 '13 at 21:15
  • You are welcome. Sorry for the inelegant writeup. Was working on one computer, lost it all, did not have the patience to polish the hurriedly typed second version. – André Nicolas Aug 18 '13 at 21:25