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Define a Thales space to be a topological space $\mathcal{X}=(X,\tau)$ such that there is some $d:X^2\rightarrow\mathbb{R}$ such that the following hold:

  • The map $d$ is a metric and the topology it induces is $\tau$.

  • The map $\mathbb{R}^2\rightarrow\mathbb{R}: (x,y)\mapsto xy$ is (first-order, with parameters) definable in the two-sorted structure $$(X\sqcup \mathbb{R}; +,<,d).$$

For example, the usual metric witnesses that $\mathbb{R}^n$ is a Thales space for each $n>1$ as a consequence of Thales' intercept theorem (hence the name). [Stupid claims removed.]

Ultimately I'd love an exact characterization of Thales spaces, but I think that's rather ambitious. An easier question would be something along the lines of, "Is there a non-pathological Thales space which doesn't lean on the intersection theorem?" I'll make that precise in the following way:

Is there a Thales space which is separable and into which $\mathbb{R}^2$ does not continuously embed?

(As Eric Wofsey's answer demonstrate separability doesn't actually rule out pathological examples, but it's what I asked.)

I'd love to add "connected" but that might make the question too hard.

Noah Schweber
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  • Incidentally, we can also ask the dual question: given a space, what operations on $\mathbb{R}$ can metrics on that space define? Specifically, the "right" version of this question seems to be the following: fixing a metrizable space $\mathcal{X}=(X,\tau)$, say that $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is metrically definable on $\mathcal{X}$ iff there are finitely many metrics $d_1,...,d_n$, each inducing $\tau$, such that $f$ is first-order definable with parameters over $(X\sqcup\mathbb{R}; +,\times,<,d_1,...,d_n)$ - the "finitely many" is to avoid issues of simultaneous definability. – Noah Schweber Dec 16 '20 at 19:34
  • But that seems extremely hard to me, even in relatively simple cases. – Noah Schweber Dec 16 '20 at 19:34
  • Can you add to this post to make it understandable for people who don't have background in logic/model theory? Or would that be too long a story? – Mike F Dec 16 '20 at 19:57
  • @MikeF That would indeed be too long a story. But here might be a good source - the question is about a particular case, but the answer is general. That answer assumes an existing understanding of "first-order sentence," "first-order structure," and the satisfaction relation $\models$. – Noah Schweber Dec 16 '20 at 20:00
  • Out of curiosity, why the downvote? – Noah Schweber Dec 31 '20 at 19:14
  • Wasn't from me! :) – Mike F Dec 31 '20 at 20:57
  • @MikeF I didn't think so, and to be honest I don't expect to hear back - but every so often I do hear back (and with a good reason even), so it doesn't hurt to ask. – Noah Schweber Dec 31 '20 at 21:35

1 Answers1

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Thales spaces have a very simple characterization: they are exactly the noncompact metrizable spaces.

First, using continuity of multiplication, to define multiplication, it suffices to be able to define multiplication on a dense subset of $\mathbb{R}$. That is, suppose there is a formula $P(x,y,z)$ such that $P(x,y,z)$ implies $xy=z$ and for all $x$ and $y$ in some dense subset of $\mathbb{R}$, there exists a $z$ such that $P(x,y,z)$. Then we can take a formula $Q(x,y,z)$ that says for all $\epsilon>0$, there exist $x',y'$, and $z'$ within $\epsilon$ of $x,y,$ and $z$ such that $P(x',y',z')$, and this will define multiplication.

Note moreover that for each $x,y\in\mathbb{R}^2$, there is a finite set $F_{x,y}$ of points in $\mathbb{R}^2$ which is enough to "witness" the value of the product $xy$: you just need the finitely many points involved in the geometric construction of the product, plus a few more finitely many points to witness that the two parallel lines really are parallel (for instance, the vertices of a rhombus with sides on the parallel lines). Thus if $H$ is any inner product space and $X$ is any subspace of $H$ that contains a subset isometric to $F_{x,y}$ for each $x,y\in\mathbb{Q}$, then $X$ can define multiplication on $\mathbb{Q}$ and thus is a Thales space by the previous paragraph.

Now suppose $X$ is a noncompact metrizable space. Then $X$ embeds in a Banach space, and every Banach space is homeomorphic to a Hilbert space, so we may assume $X$ is a subspace of a Hilbert space $H$ of dimension at least $2$. Take a sequence $(x_n)$ of distinct points in $X$ with no accumulation point. Then there is a continuous function $f:X\to\mathbb{R}$ such that $f(x_n)\to\infty$. Replacing $H$ with $H\times\mathbb{R}$ and using $f$ on the second coordinate, we may assume that $\|x_n\|\to\infty$ in $H$. Choosing $f$ to grow fast enough on $(x_n)$, we may moreover assume that the norms $\|x_n\|$ are nonzero and strictly increasing with $n$.

Now take a sequence of isometric copies of $F_{x,y}$ in $H$ for each $x,y\in\mathbb{Q}$ which are eventually outside of any given ball. We may moreover arrange that all the points in these copies have distinct norms. Thus, we can get a sequence $(y_n)$ in $H$ with the norms $\|y_n\|$ nonzero and strictly increasing to $\infty$ such that any subspace of $H$ which contains every $y_n$ is a Thales space.

Finally, I claim there is a homeomorphism $H\to H$ that maps $x_n$ to $y_n$ for each $n$, so that $X$ is homeomorphic to a subspace of $H$ that contains each $y_n$ and thus is a Thales space. You can construct such a homeomorphism, for instance, using polar coordinates. On the radial coordinate, take a homeomorphism $[0,\infty)\to [0,\infty)$ which maps $\|x_n\|$ to $\|y_n\|$. On the sphere of radius $\|x_n\|$, use a rotation of the unit sphere in $H$ that maps $x_n/\|x_n\|$ to $y_n/\|y_n\|$, and interpolate between these rotations on the other spheres.

Thus, every noncompact metrizable space is a Thales space. Conversely, I claim multiplication can never be definable from a bounded metric, and so a compact space is never a Thales space. Indeed, suppose $d$ is a metric on $X$ and $N$ is an integer such that $d(x,y)<N$ for all $x,y\in X$. Note that by splitting each real number as the sum of an element of $N\mathbb{Z}$ and and element of $[0,N)$, the structure $(X\sqcup \mathbb{R}; +,<,d)$ is interpretable in the three-sorted structure $$(X\sqcup [0,N)\sqcup N\mathbb{Z};+_{[0,N)},<_{[0,N)},+_{N\mathbb{Z}},<_{N\mathbb{Z}},d_{[0,N)})$$ where $+_{[0,N)}$ is addition mod $N$ on $[0,N)$, $<_{[0,N)},+_{N\mathbb{Z}},$ and $<_{N\mathbb{Z}}$ are the obvious restrictions, and $d_{[0,N)}$ is the metric considered as a map $X^2\to [0,N)$. But in this structure, the sort $N\mathbb{Z}$ does not interact at all with the other two sorts. As a result, if multiplication were definable in $(X\sqcup \mathbb{R}; +,<,d)$, then multiplication would be definable in $(N\mathbb{Z};+,<)$. But multiplication is not definable in $(N\mathbb{Z};+,<)$ (for instance, if it were, it would follow easily that multiplication is definable in $(\mathbb{N},+)$).

Eric Wofsey
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  • I'm now interested in closing the gap ... – Noah Schweber Dec 17 '20 at 02:56
  • See my most recent edit. :) – Eric Wofsey Dec 17 '20 at 04:58
  • Delightful! I definitely didn't guess that this would be the answer. – Noah Schweber Dec 17 '20 at 05:21
  • Is there an obvious way to extend this to (say) "Every continuous $\mathbb{R}\rightarrow\mathbb{R}$ is definable via some metric on $\mathbb{R}^2$?" At a glance I don't see it, since the details of Thales' construction actually seem to matter a bit, but I don't see any particular obstacles either. – Noah Schweber Dec 17 '20 at 07:21
  • I haven't thought through the details but I bet you could do something like this. Take two different definable ways a metric can behave locally (for instance, maybe you can define "the metric is Euclidean on the $\epsilon$-ball around $x$", and maybe you can also do it with something like the $\ell^1$ metric). Now consider the following statement $P(q,r)$: "there exist points $x$ and $y$ such that the $q$-ball around $x$ has Behavior 1, and the $r$-ball around $y$ has Behavior 2, and $d(x,y)=q+r$". – Eric Wofsey Dec 17 '20 at 15:06
  • For each $q\in\mathbb{Q}^+$ you can arrange for $P(q,r)$ to be true for some given $r$. By making these instances far away from each other and making the metric highly irregular in between them, you can arrange that $P(q,r)$ is true in no other cases. In thiis way you can define an arbitrary function $\mathbb{Q}^+\to\mathbb{R}^+$. If it happens to extend continuously to $\mathbb{R}^+$, you can then extend the definition using continuity. – Eric Wofsey Dec 17 '20 at 15:08
  • I'm not sure I see it. Fixing some "nice" $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$, suppose we build an appropriate $P$. We can certainly define a function $g$ such that $f\upharpoonright\mathbb{Q}^+=g\upharpoonright\mathbb{Q}^+$, but in order to get $f$ from $g$ it seems to me we need to know $\mathbb{Q}^+$ as well. For example, the naive definition of $g$ is "$g(a)=b$ if $b$ is the unique number satisfying $P(a,b)$, and (say) $g(a)=0$ if no such $b$ exists." Regardless of $f$, this $g$ could be highly discontinuous due to its behavior off $\mathbb{Q}^+$ and I don't see how to get around that. – Noah Schweber Dec 18 '20 at 06:13
  • That step is basically the same as the second paragraph of my answer. – Eric Wofsey Dec 18 '20 at 06:21
  • Another way to put it is that we do have $\mathbb{Q}^+$: we can define it as the set of $a$ such that there exists $b$ such that $P(a,b)$. – Eric Wofsey Dec 18 '20 at 06:23