Topology Puzzles

November 11, 2017

Posted by John Baez

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Let’s say the closed unit interval $[0,1]$ maps onto a metric space $X$ if there is a continuous map from $[0,1]$ onto $X$ . Similarly for the Cantor set.

Puzzle 0. Does the Cantor set map onto the closed unit interval, and/or vice versa?

Puzzle 1. Which metric spaces does the closed unit interval map onto?

Puzzle 2. Which metric spaces does the Cantor set map onto?

The first one is easy; the second two are well-known… but still, perhaps, not well-known enough!

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The answers to Puzzles 1 and 2 can be seen as ‘versal’ properties of the closed unit interval and Cantor set — like universal properties, but without the uniqueness clause.

Posted at November 11, 2017 4:38 AM UTC

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Re: Topology Puzzles

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Please forgive me for asking, but what is the point of posing these puzzles? The answers are indeed widely known, and easily googled.

Posted by: Todd Trimble on November 11, 2017 4:13 PM | Permalink | Reply to this

Re: Topology Puzzles

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I just learned these facts and think they’re cool. If someone who doesn’t know them yet finds out these facts by googling, I’ve succeeded: I’ve gotten someone to learn something cool! And if someone figures out the answers without ‘cheating’ in that way, it’s even better.

I was also trying to start a conversation. These questions raise other questions in my mind?

Puzzle 3. Which spaces $X$ have easily stated ‘versal’ properties of the form “all spaces of type $T$ are mapped onto by $X$ ”?

That’s probably a bit too broad to focus a discussion. Versal properties are weaker than universal properties, in the sense that there may be non-isomorphic spaces with the same versal property. For example, it’s pretty easy to see that $[0,1]^2$ maps onto the same metric spaces as $[0,1]$ . So one could ask:

Puzzle 4. Which metric spaces map onto exactly the same metric spaces that $[0,1]$ maps onto?

Posted by: John Baez on November 11, 2017 6:01 PM | Permalink | Reply to this

Re: Topology Puzzles

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Okay, I didn’t know whether these puzzles were stepping stones to something else you wanted to talk about, or what.

(Actually, it feels like calling them “puzzles” is a little strange, or at least so for Puzzle 1, since AFAIK the characterization is a pretty difficult theorem that I wouldn’t expect someone to work out to completion without considerable effort, if they didn’t know it already. Usually I think of a puzzle as being something lighter. Puzzle 2 probably wouldn’t require as much mental labor to write out a proof for.)

Some time ago I asked on MathOverflow the situation for $\mathbb{R}$ instead of $I = [0, 1]$ , and I got an answer (which piggybacks on the answer to Puzzle 1), but I didn’t find the answer especially exciting.

Posted by: Todd Trimble on November 11, 2017 6:12 PM | Permalink | Reply to this

Re: Topology Puzzles

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Todd wrote:

Okay, I didn’t know whether these puzzles were stepping stones to something else you wanted to talk about, or what.

No big agenda. These days I’m too busy working on math to think about math ‘just for fun’ the way I used to. But I just ran into a couple of fun facts and felt like mentioning them.

Actually, it feels like calling them “puzzles” is a little strange […]

True. It’s kind of a joke: over on Google+, I post a lot of ‘puzzles’ of widely ranging difficulty. Some I know the answers to, some I wish I knew the answer to but don’t. Some require only basic skills like counting, just to get people involved in looking at pictures more actively. Some are much harder. They’re all just ways of getting conversations started.

For example, there’s a guy with a flat Earth theory who thinks gravity is caused by the upward acceleration of the Earth. He actually knows about special relativity, so I posed this:

Puzzle. Assume the Earth is 6000 years old and has been accelerating at 9.8 meters per second per second throughout this time, as measured in its own instantaneous rest frame. If it started at rest, how fast is it going now? Use the formula for Lorentz contractions to compute the thickness of the Earth today as measured by an observer at rest.

Or:

Puzzle. When we fold a regular dodecahedron back to a cube as below, does it fit together snugly, or is there some empty space left? What fraction of the cube is filled?

Or:

Puzzle. Suppose two red supergiants containing neutron stars collide. What happens then?

Unfortunately I didn’t have enough time to write a spiel leading up to the puzzles this time. Let me try to make up for that now:

We can define a preorder on nonempty compact metric spaces where $X \ge Y$ if we can map $X$ onto $Y$ . (This is a cheap trick for turning a versal property into a universal property.)

Since ordinary people like partial orders more than preorders, let’s say two nonempty compact metric spaces $X$ and $Y$ are equivalent if $X \le Y$ and $X \ge Y$ . Let $CompMet$ be the collection of equivalence classes. This gets a partial ordering where $[X] \ge [Y]$ iff $X \ge Y$ .

Puzzle 5. Say as much as you can about the poset $CompMet$ .

For example, the answer to Puzzle 3 says the Cantor set sits at the very top of this partial ordering.

(I had to exclude the empty set for this to be true! Sorry, empty set!)

Also, there’s no way a compact metric space with at most $n$ connected components can map onto one with more than $n$ connected components. So, if we let $CompMet_n$ be the collection of equivalence classes of compact metric spaces with $n$ connected components, we can’t have anyone in $CompMet_n$ that’s $\ge$ anyone in $CompMet_{m}$ if $n < m$ .

(Here $m = 1,2,3,\dots, \aleph_0$ , since there can be at most countably many connected components in a compact metric space.)

So, $CompMet$ has a kind of ‘grading’ by the number of connected components, and it’s probably enough to focus in on $CompMet_1$ .

Puzzle 6. Does $CompMet_1$ have a top element?

But $CompNet$ has another grading, by the number of path-connected components. The number of path-connected components is always $\ge$ the number of connected components.

There’s more structure, too, some of which is exhibited in Puzzle 2.

Posted by: John Baez on November 11, 2017 9:48 PM | Permalink | Reply to this

Re: Topology Puzzles

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I haven’t looked carefully into your Puzzle 5 yet, but it looks interesting!

Just one little thing:

(Here $m = 1,2,3,\dots, \aleph_0$ , since there can be at most countably many connected components in a compact metric space.)

Careful! Cantor space is totally disconnected, i.e. each point is its own connected component, so there are $2^{\aleph_0}$ connected components there.

Posted by: Todd Trimble on November 11, 2017 10:47 PM | Permalink | Reply to this

Re: Topology Puzzles

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Whoops! I had another fact on my mind: every compact metric space has a dense subset of cardinality at most $\aleph_0$ . But that’s not very relevant here.

Putting everything together, we can see that $2^{\aleph_0}$ is the largest number of connected components a compact metric space can have.

Posted by: John Baez on November 11, 2017 11:09 PM | Permalink | Reply to this

Re: Topology Puzzles

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I posed Puzzle 5 over on MathOverflow and was pleased to see that this puzzle was apparently asked by Hahn, and answered by Waraszkiewicz in 1934.

I wrote:

Fact 1. The Cantor set $K$ is “universal” among nonempty compact metric spaces in the following sense: given any nonempty compact metric space $X$ , there exists a continuous surjection $f\colon K \to X$ .

Fact 2. The closed interval $I$ has a similar “universal” property among nonempty compact connected and locally connected metric spaces: given any such space $X$ , there exists a continuous surjection $f \colon I \to X$ .

This makes me wonder: is there a compact connected metric space $J$ such that for any nonempty compact connected metric space $X$ , there exists a continuous surjection $f \colon J \to X$ ?

Such a space $J$ , if it exists, would be ‘intermediate’ between $I$ and $K$ : there would need to be continuous surjections

$K \to J \to I$

Fact 1 is sometimes called the Alexandroff–Hausdorff theorem, since appeared in the second edition of Felix Hausdorff’s Mengenlehre in 1927 and also in an article by Pavel Alexandroff published in Mathematischen Annalen in the same year. Fact 2 was proved by Hans Hahn in 1914 and reproved by him more nicely in 1928. For a nice history of these results, see:

L. Koudela, The Hausdorff–Alexandroff theorem and its application in theory of curves, WDS’07 Proceedings of Contributed Papers, Part I, 2007, pp. 257–260.

One may rightly complain that “universal” is the wrong word above, since we’re not claiming there exists a unique continuous surjection, and indeed there’s usually not. A better term is versal. There can be two non-homeomorphic spaces having the same versal property. For example, $I^2$ would work just as well as $I$ in Fact 2, thanks to the existence of space-filling curves.

Nonetheless we can create a category in which these versal properties become universal, by a cheap trick. Let $\mathrm{CompMet}$ be the collection of all homeomorphism classes of nonempty compact metric spaces, and put a partial order on this where $[X] \ge [Y]$ iff there exists a continuous surjection $f \colon X \to Y$ . The homeomorphism class of the Cantor set is the top element of the poset $\mathrm{CompMet}$ . My question asks if the subset of $\mathrm{CompMet}$ coming from connected compact metric spaces has a top element.

I’d appreciate any interesting information on this poset $\mathrm{CompMet}$ .

For example, I think that there’s a map sending each element of $\mathrm{CompNet}$ to its number of connected components, and I think that this is an order-preserving map from $\mathrm{CompMet}$ to the cardinals less than or equal to the continuum. But there also seems to be an order-preserving map sending each element of $\mathrm{CompNet}$ to its number of path-connected components. Are there other interesting maps like this?

Jeremy Brazas said that no such space $J$ exists:

There is no such continuum. See

Z. Waraszkiewicz, Sur un problème de M. H. Hahn, Fund. Math. 22 (1934) 180–205.

Waraszkiewicz constructed an uncountable family $W$ of continua in the plane called Waraszkiewicz spirals so that no continuum can be mapped continuously onto every continuum in $W$ . Start with the space $X=S^1\cup [1,2]$ and consider the maps $f,g:X\to X$ where

$f(x)=x$ if $x\in S^1$ , $f(x)=\exp(4\pi x)$ for $1\leq x\leq \frac{3}{2}$ , and $f(x)=2(x-1)$ for $\frac{3}{2}\leq x\leq 2$

$g(x)=x$ if $x\in S^1$ , $g(x)=\exp(-4\pi x)$ for $1\leq x\leq \frac{3}{2}$ , and $g(x)=2(x-1)$ for $\frac{3}{2}\leq x\leq 2$

Now $W$ is the collection of all inverse limits of all inverse sequences $(X_i,h_i)$ where $X_i=X$ for all $i$ and $h_i\in\{f,g\}$ .

For a short and readable description (where I found the construction) see:

W. T. Ingram, Concerning images of continua, Topology Proceedings 16 (1991) 89–93.

This non-existence result has been improved upon a few times, for instance in the following:

S. B. Nadler, The nonexistence of almost continuous surjections between certain continua, Topology and its Applications, 154 (2007) 1008–1014.

Posted by: John Baez on November 12, 2017 6:44 AM | Permalink | Reply to this

Re: Topology Puzzles

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Todd,

I think John posted these here because I asked a very vague and ill-formed question about how to view the answer to puzzle 3 in more generalised and constructive topological settings. Is it reasonable to consider the class of spaces in a constructive/synthetic theory of spaces that are mapped onto by the Cantor set (concretely, say, $\{0,1\}^\omega$ ), rather than the usual constructivised version of the answer?

Posted by: David Roberts on November 12, 2017 3:28 AM | Permalink | Reply to this

Re: Topology Puzzles

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Yes, your post triggered mine… but I mainly got excited because you mentioned that the Cantor set maps onto every nonempty compact metric space… and when I looked up a reference, I discovered that the closed interval has a similar but more complicated ‘versal’ property. This got me interested in understanding all such versal properties… which amounts to understanding the poset $\mathrm{CompMet}$ mentioned above.

Posted by: John Baez on November 12, 2017 6:35 AM | Permalink | Reply to this

Re: Topology Puzzles

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David,

I must have missed it. Can you point to where you discussed this, please?

Posted by: Todd Trimble on November 12, 2017 1:25 PM | Permalink | Reply to this

Re: Topology Puzzles

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Here: https://plus.google.com/+DavidRoberts/posts/2yTY59uFA6d. But I warn you, it was not well thought out. I was being hasty and also not wanting to explain myself fully to leave space for discussion.

Posted by: David Roberts on November 13, 2017 12:56 AM | Permalink | Reply to this

Re: Topology Puzzles

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Since the cat is out of the bag, maybe we could talk about Puzzle 3 a little, with an eye to constructive aspects. I’m not well-trained in constructive mathematics, so there’s a good chance I’ll learn something here.

I would guess that if you can prove “constructively” (whatever you take that to mean) that a compact metric space $(X, d)$ has a countable dense subset $x_n$ , then the coast is clear. Assume, as we may, that the metric $d$ is valued in $I = [0, 1]$ . The trick is first to use that sequence to embed $X$ in the Hilbert cube $I^\mathbb{N}$ , by

$u: X \hookrightarrow I^\mathbb{N}$

$x \mapsto (d(x_n, x))_{n \in \mathbb{N}}.$

(By the way, this says that the Hilbert cube $I^\mathbb{N}$ also enjoys a ‘versal’ property: any compact metric space embeds into it.)

Now, Cantor space $C = \{0, 1\}^\mathbb{N}$ maps onto $I$ (Puzzle 1) by sending a sequence $(a_n)$ of $0$ ’s and $1$ ’s to $\sum_{n \geq 0} \frac{a_n}{2^{n+1}}$ . Call this map $\lambda$ (partly in honor of Lebesgue: this map is closely related to the Cantor-Lebesgue function). Then there is a continuous surjection of $C$ onto the Hilbert cube, via a series of maps

$C = \{0, 1\}^\mathbb{N} \cong \{0, 1\}^{\mathbb{N} \times \mathbb{N}} \cong (\{0, 1\}^\mathbb{N})^\mathbb{N} \stackrel{\lambda^\mathbb{N}}{\to} I^\mathbb{N}.$

By taking a pullback in $Top$ , the category of topological spaces,

$\array{ K & \hookrightarrow & C \\ \downarrow \mathrlap{q} & & \downarrow \mathrlap{surj} \\ X & \stackrel{u}{\hookrightarrow} & I^\mathbb{N} }$

we get a continuous surjection $q: K \twoheadrightarrow X$ of a subspace $K$ of $C$ onto $X$ .

Now all we need is a continuous surjection $r: C \to K$ , since then $p = q \circ r: C \to X$ is the desired surjection. But I claim the inclusion $K \hookrightarrow C$ has a retraction $r$ , constructively. Simply identify $C$ with the subspace of numbers in $[0, 1]$ which, written in base $5$ , have only $0$ ’s and $4$ ’s. Then the average $\frac{x+y}{2}$ of any two $x, y \in C$ does not belong to $C$ (consider the first place where the base $5$ expansion of $x, y$ differ; then the average will have a $2$ in that place). It follows that for any $x \in C$ , there is a unique $r(x) \in K$ such that

$d(x, r(x)) = \inf \{d(x, y): y \in K\}$

and $x \mapsto r(x)$ provides the desired retraction $r$ .

Posted by: Todd Trimble on November 12, 2017 2:42 PM | Permalink | Reply to this

Re: Topology Puzzles

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Then the average $\frac{x+y}{2}$ of any two $x, y \in C$ does not belong to $C$

I think you mean the average of any two distinct $x,y\in C$ does not belong to $C$ . This smells like somewhere that you may get into trouble constructively, since $C$ doesn’t have decidable equality.

It follows that for any $x \in C$ , there is a unique $r(x) \in K$ such that $d(x,r(x))=\inf\{d(x,y):y\in K\}$ .

I lost you completely here; how does it follow?

Posted by: Mike Shulman on November 12, 2017 10:03 PM | Permalink | Reply to this

Re: Topology Puzzles

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I did mean “distinct” $x, y \in C$ ; good catch. So some repair work would probably have to be done there.

As for your question: if there were two distinct points $x', x'' \in K$ such that $d(x, x') = d(x, x'')$ realize the minimum distance from $x$ to the (compact) set $K$ , then $x$ would be the midpoint between $x'$ and $x''$ . But that midpoint (i.e. average) cannot lie in $C$ .

Posted by: Todd Trimble on November 12, 2017 11:27 PM | Permalink | Reply to this

Re: Topology Puzzles

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Actually now I think there is no problem constructively, if you interpret “distinct” to mean apart. On the one hand, to conclude “the average of two distinct $x,y\in C$ does not belong to $C$ ” you need $x$ and $y$ to be apart, since that’s what gives you the existence of a decimal place where they differ. On the other hand, to show that $x'=x''$ it suffices to show that $x'$ and $x''$ are not apart, since the apartness on a (Hausdorff) metric space is tight.

How do you know that there exists at least one $r(x)\in K$ ?

Posted by: Mike Shulman on November 13, 2017 3:09 AM | Permalink | Reply to this

Re: Topology Puzzles

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I know there’s at least such $r(x)$ because for a given $x \in C$ , the function $d(x, -): K \to [0, 1]$ attains a minimum by compactness of $K$ .

Posted by: Todd Trimble on November 13, 2017 11:45 AM | Permalink | Reply to this

Re: Topology Puzzles

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Okay, that’s somewhere that constructivity may bite us: in general, continuous functions on compact sets don’t necessarily achieve extrema.

Posted by: Mike Shulman on November 13, 2017 9:30 PM | Permalink | Reply to this

Re: Topology Puzzles

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Possibly so. I was wondering a little whether the unique existence of the construction could be played to advantage. (Also, what does locale theory say about extreme value theorems for compact regular locales? The nLab is a little vague on this matter.)

Posted by: Todd Trimble on November 13, 2017 11:23 PM | Permalink | Reply to this

Re: Topology Puzzles

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Possibly. I know less about constructive versions of the extreme value theorem than I do about the intermediate value theorem. It’s true that at least in the simple topological counterexamples, the extreme value cannot be asserted to be unique. But proofs of the EVT generally proceed through approximations, so it’s unclear to me how to make use of a uniqueness for the exact extremum. Can we make any sort of uniqueness statement that would apply to approximations as well?

Posted by: Mike Shulman on November 14, 2017 5:37 AM | Permalink | Reply to this

Re: Topology Puzzles

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I don’t know if this will work, but one could observe that in this situation, the function $d(x, -): K \to [0, 1]$ for $x \in C$ is a bijective Lipschitz $1$ map, and so I guess takes $K$ to a complete totally bounded subset of $[0, 1]$ . If it is constructively true that such have minima, then I would guess that this argument does work after all.

Meanwhile however I have located a source that Cantor space is versal, constructively (see my comment here).

Posted by: Todd Trimble on November 16, 2017 2:33 AM | Permalink | Reply to this

Re: Topology Puzzles

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Well, while reading around a little, I did find this statement on the constructive versal property of Cantor space $\mathbf{2}^\mathbb{N}$ in an article here, page 3:

Theorem: If $X$ is an inhabited, complete, and totally bounded metric space, then there exists a uniformly continuous, open mapping of $\mathbf{2}^\mathbb{N}$ onto $X$ .

(My readings furthermore suggest that “complete + totally bounded” is a preferred way of formulating the notion of compact metric space for constructivists.)

So, assuming the statement is correct, at least we know there is a constructivist version of Puzzle 2 (which earlier I preferred to as Puzzle 3, forgetting that John began his enumeration at $0$ ).

I haven’t looked at the cited source for this statement to see how they go about proving it.

Posted by: Todd Trimble on November 16, 2017 2:22 AM | Permalink | Reply to this

Re: Topology Puzzles

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I don’t have that reference to hand right now, but I do have a proof of the same result in Troelstra and van Dalen’s Constructivism in Mathematics (Volume 2, Chapter 7). As I expected, it uses countable choice, which lots of constructivists do; so I guess probably the answer depends on what you take “constructive” to mean. Troelstra and van Dalen suggest that in the absence of countable choice one ought to essentially define “complete totally bounded metric space” so as to make this theorem true.

I wonder whether we could do better by regarding $X$ as a compact metric locale, and likewise Cantor space as a locale? Constructing a surjection from something like Cantor space onto an arbitrary set seems sure to use some kind of choice, whereas a “surjection” (i.e. epimorphism) of locales need not, I think, be set-theoretically surjective on points.

Posted by: Mike Shulman on November 16, 2017 5:30 AM | Permalink | Reply to this

Re: Topology Puzzles

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one ought to essentially define “complete totally bounded metric space” so as to make this theorem true.

aha, that’s what I was wondering when I wrote my original G+ post. Thanks for digging this out.

I agree that a localic version would be even better. The nLab has scant details at the page ‘Cantor space’. It seems the fan theorem is equivalent to the Cantor locale having enough points (due to Michael P. Fourman and Robin J. Grayson in the 1982 paper “Formal spaces.”, I think).

Posted by: David Roberts on November 16, 2017 11:44 AM | Permalink | Reply to this

Re: Topology Puzzles

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I bumped into another interesting compact metric space with a nice ‘versal property’: the pseudoarc.

The pseudo-arc can be described a subset of the plane, but it’s tough to visualize: Wayne Lewis and Piotr Minc’s paper Drawing the pseudo-arc gives a “somewhat disappointing” picture of the pseudo-arc which looks like an ugly black square with ragged edges, and then proves some theorems explaining why it’s hard to do better.

The pseudo-arc is a chainable continuum:

Recall that a continuum is a compact connected metric space. For a compact metric space $X$ , we say that an open cover $\mathcal{U}$ of $X$ refines an open cover $\mathcal{V}$ if each element of $\mathcal{U}$ is contained in an element of $\mathcal{V}$ . We call a continuum $X$ chainable if each open cover of $X$ is refined by an open cover $U_1, \dots, U_n$ such that for $i, j \le n$ , $U_i \cap U_j \ne \emptyset$ if and only if $|i - j| \le 1$ . Such a cover of X is called a chain.

Moreover, in 1962 Mioduszewski proved that the pseudo-arc maps onto every chainable continuum!

The pseudo-arc has other amazing properties:

A continuum is indecomposable if it is not the union of two proper subcontinua. A continuum is hereditarily indecomposable if each of its subcontinua is indecomposable. The pseudoarc is the unique hereditarily indecomposable chainable continuum.

But the pseudo-arc is also amazingly common among continua:

It is also the generic continuum: in the (compact) space of all subcontinua of $[0,1]^{\mathbb{N}}$ equipped with the Hausdorff metric, homeomorphic copies of the pseudo-arc form a dense $G_\delta$ set.

These quotes are from Trevor L. Irwin and Salawomir Solecki’s paper Projective Fraisse limits and the pseudo-arc.

Posted by: John Baez on November 18, 2017 7:38 AM | Permalink | Reply to this

Re: Topology Puzzles

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Thanks for mentioning this. We have some material on pseudo-arcs at the nLab, where it is observed that they often arise as inverse limits of diagrams of type

$\ldots \stackrel{f}{\to} I \stackrel{f}{\to} I \stackrel{f}{\to} I$

where $f: I \to I$ “doubles back” on itself a bunch of times (see Example 4 here, page 38 of 58). But the material on chainable continua deserves to be mentioned in the nLab as well.

Posted by: Todd Trimble on November 18, 2017 4:14 PM | Permalink | Reply to this

Re: Topology Puzzles

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For more fun with the pseudo-arc, see my post on Azimuth:

• A universal snake-like continuum.

Posted by: John Baez on November 27, 2017 1:00 AM | Permalink | Reply to this

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