No surjection onto function space for manifold X

[Note: highly tech­ni­cal. Skip if topol­ogy is not your thing]

In his post on for­mal open prob­lems in de­ci­sion the­ory, 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 .

I thought I had an ar­gu­ment for how could be such an . But that ar­gu­ment is wrong, as I’ll demon­strate in this post. In­stead I will show that:

  • 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.

Proof

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 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.

It will first as­sume that is Fréchet–Urysohn, -com­pact, and a space. Once the proof is com­pleted, it will then go back and re­move the con­di­tion.

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.

So put the dis­crete sets aside. Since is not ex­clu­sively made up of iso­lated points, it must con­tain a limit point; call it . Since is Fréchet–Urysohn, there ex­ists a se­quence , , that con­verges on .

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

We aim to show, us­ing As­coli’s The­o­rem, that any com­pact sub­set of has empty in­te­rior—thus con­tains no open sets.

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

By the Baire cat­e­gory the­o­rem, if is a com­plete met­ric space, then it is a Baire space.

Since is com­pact, the com­pact-open and the uniform con­ver­gence topol­ogy match up on . So has a uniform norm com­pat­i­ble with its topol­ogy: .

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.

Be­cause is , then so is . But any Lin­delöf reg­u­lar space is nor­mal (see the­o­rem 14), hence, since is also , then it must be Haus­dorff nor­mal, ie .

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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