No surjection onto function space for manifold Xpost by Stuart_Armstrong · 2019-01-09T18:07:26.157Z · score: 22 (6 votes) · LW · GW · 0 comments
Proof A note on topologies The steps of the proof Excluding discrete spaces, and finding compact Y⊂X The restriction map C(X,I)→C(Y,I) is continuous surjective Compact subsets of C(Y,I) have empty interior C(Y,I) is a Baire space There is no surjective continuous map s:X→C(X,I) Removing the T4 condition No comments
[Note: highly technical. Skip if topology is not your thing]
In his post on formal open problems in decision theory [LW · GW], Scott asked whether there could exist a topological space and a continuous surjection from to . Here, is the closed unit interval and is the set of continuous functions from to .
- Let be a manifold (with or without boundary), or a union of finitely or countably many manifolds. Then there is no continuous surjective map from to .
By "union", imagine the manifolds lying inside Euclidean space (or, more generically, inside a metric space), not necessarily disjoint, and taking their unions there. Note that there are many examples of such s - for example, the rationals within the reals (being countable unions of points, which are trivial manifolds).
In fact, I will show the more general:
To see that this more general result implies the one above, note that manifolds are -compact, and that if is -compact, it can be covered by countably many compact sets, so can be covered by countably many sets of countably many compact sets, which is just countably many. Finally, all metric spaces are Fréchet–Urysohn.
Fréchet–Urysohn basically means "convergence of subsequences makes sense in the topology", and is not a strong restriction; indeed all first-countable spaces are Fréchet–Urysohn.
A note on topologies
Now is well-defined as a set, but it needs a topology to discuss issues of continuity. There are three natural topologies on it: the topology of uniform convergence, the compact-open topology (which, on this set, is equal to the topology of compact convergence), and the topology of pointwise convergence.
The one most people use, and that I'll be using, is the compact-open topology.
These are refinements of each other as so:
- uniform convergence open-compact pointwise convergence.
Thus, if we could find a surjective in the uniform convergence topology, it would still be continuous in the compact-open topology. Conversely, if we could show that no such maps exist in the pointwise convergence topology, there would also be no maps in the compact-open one. Unfortunately, the partial results I've found are opposite: I believe I can show that no maps exist for general 's for uniform convergence, and I have a vague argument [LW · GW] that may allow us to construct one in the pointwise convergence topology. Neither of these help here.
Anyway, onwards and upwards!
The steps of the proof
The proof will have lots of technical terminology (with links) but will be short on detailed explanations of this terminology.
Why do it in that order? Because without the condition, it is a proof by contradiction, assuming a continuous surjection to do most of the work. I have nothing against proofs by contradiction, but, if ever anyone wants to refine or extend my proof, I want to make clear which results are really true for , and which ones only arise via the contradiction.
The proof will proceed by these steps:
- Excluding discrete spaces, and finding compact .
- The restriction map is continuous surjective.
- Compact subsets of have empty interior.
- is a Baire space.
- There is no surjective continuous map .
- Removing the condition.
Excluding discrete spaces, and finding compact
If is discrete, then , which has strictly higher cardinality than , so surjective maps are impossible at the set level.
Then define ; it's clear that it is compact in the subspace topology. Since is also , must be closed in .
The restriction map is continuous surjective
Any function is a map by restriction, so there is a map .
Since is and is closed, for any continuous function , there exists, by the Tietze extension theorem, a continuous such that for all . So is surjective.
We now want to show that is also continuous.
A subbase for consists of all where is compact in and is open, and iff .
Note that since has the subspace topology in , these must be compact in as well. So define , consisting of continuous functions from to with . Then , and thus the pre-image of the sets of this subbasis is open.
Since it suffices to check continuity on a subbasis, this shows that itself is continuous.
Compact subsets of have empty interior
The subbasis for the topology on consists of sets of the form , where is compact in , is open in , and iff .
The compact subsets of are a) finite collections of points , , b) collections of points that include all points for , along with , and c) itself.
Let be an intersection of finitely many . If is non-empy, it is defined by an and a collection of sets open in . Then belongs to iff:
Since is non-empty and is continuous, we must have and having non-zero intersection. This actually implies that and have non-zero intersection.
Note that this includes all possible , as we can always set (or or ), thus removing that particular restriction.
We want to show that such an cannot be equicontinuous at .
To do so, fix points for , and , with
Then consider the functions , for , defined by:
- for ,
- for ,
- for .
These are all in , and continuous (since is discrete except at ). If were equicontinuous then for all , there would exists an open set in containing such that for all and all .
Since is open, there exists a , . Set ; then .
Since is not equicontinuous, and is both compact and Hausdorff (and hence pre-compact Hausdorff), cannot be contained in a compact subset of , by Ascoli's Theorem.
Any open set in consists of unions of sets of the type , so no (non-empty) open set is contained in compact subset of .
is a Baire space
So is thus a metric space. Since is complete, the uniform limit theorem shows is a complete metric space (and hence a Baire space).
There is no surjective continuous map
Let be any continuous map.
Since is -compact, for compact . The set is compact in , since the continuous image of compact spaces are compact.
Let . Since compact sets in must have empty interior, the complement of has empty interior. Therefore the are dense in .
The space is equal to:
- , where denotes taking the complement in .
But, since is a Baire space, is dense, hence certainly not zero, and so .
Since is surjective , the cannot be surjective onto , proving the result.
Removing the condition
We only used to prove the properties of the subsequence . Without , we can use the continuous surjection itself. So now assume that such an exists, and we shall try and find a contradiction.
The -compact property implies is Lindelöf. Lindelöf is preserved by continuous maps, so because of , is also Lindelöf.
Now contains at least the constant functions, so must be of uncountable cardinality. Since is a countable union of spaces, there exists a such that contains a set with infinitely many points. Pull these infinitely many points back to , using the axiom of choice to choose a set such that for each , there is a unique with .
Now, is compact, which means that it is countably compact. Now, Fréchet–Urysohn is hereditary, meaning that it applies to subspaces as well, so is Fréchet–Urysohn, which means that it is sequential. For sequential spaces, countably compact implies sequentially compact (see theorem 10).
Thus we can pick a sequence in , and, passing to a subsequence if necessary, we can find a sequence converging to a point that is not equal to any of the . Define as before.
The set must also be a convergent subsequence.
Let , then (note that is well defined here, as it is a bijection between and its image). Because is itself , there exists a function that is equal to on . Then the function is equal to on , and is an element of . Thus the restriction map is still surjective, and the topology on (indeed on any sequence tending to a point) is the same as before. The argument for being continuous goes through as before.
Then the rest of the proof proceeds as above.
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