No surjection onto function space for manifold X

[Stuart_Armstrong]

[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 $X$ and a continuous surjection $s$ from $X$ to $C(X,I)$. Here, $I$ is the closed unit interval $[0,1]$ and $C(X,I)$ is the set of continuous functions from $X$ to $I$.

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

• Let $X$ be a manifold (with or without boundary), or a union of finitely or countably many manifolds. Then there is no continuous surjective map from $X$ to $C(X,I)$.

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 $X$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:

• If $X$ is Fréchet–Urysohn, and $\sigma$-compact, then there is no continuous surjective map from $X$ to $C(X,I)$.

To see that this more general result implies the one above, note that manifolds are $\sigma$-compact, and that if $X_i$ is $\sigma$-compact, it can be covered by countably many compact sets, so ${\bigcup }_{i\in N}{X}_i$ 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 $C(X,I)$ 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 $\supset$ open-compact $\supset$ pointwise convergence.

Thus, if we could find a surjective $s:X\to C(X,I)$ 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 $T_4$ $X$’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 $X$ is Fréchet–Urysohn, $\sigma$-compact, and a $T_4$ space. Once the proof is completed, it will then go back and remove the $T_4$ condition.

Why do it in that order? Because without the $T_4$ condition, it is a proof by contradiction, assuming a continuous surjection $s:X\to C(X,I)$ 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 $X$, and which ones only arise via the contradiction.

The proof will proceed by these steps:

• Excluding discrete spaces, and finding compact $Y\subset X$.

• The restriction map $C(X,I)\to C(Y,I)$ 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 $s:X\to C(X,I)$.

• Removing the $T_4$ condition.

Excluding discrete spaces, and finding compact $Y\subset X$

If $X$ is discrete, then $C(X,I)=I^X>2^X$, which has strictly higher cardinality than $X$, so surjective maps $X\to C(X,I)$ are impossible at the set level.

So put the discrete sets aside. Since $X$ is not exclusively made up of isolated points, it must contain a limit point; call it $y_{\infty }$. Since $X$ is Fréchet–Urysohn, there exists a sequence $\{y_i{\}}_{i\in N}$, $y_i\ne y_{\infty }$, that converges on $y_{\infty }$.

Then define $Y=\{y_i{\}}_{i\in N}\cup \{y_{\infty }\}$; it’s clear that it is compact in the subspace topology. Since $X$ is also $T_2$, $Y$ must be closed in $X$.

The restriction map $C(X,I)\to C(Y,I)$ is continuous surjective

Any function $f:X\to I$ is a map $f:Y\to I$ by restriction, so there is a map $r:C(X,I)\to C(Y,I)$.

Since $X$ is $T_4$ and $Y$ is closed, for any continuous function $f:Y\to I$, there exists, by the Tietze extension theorem, a continuous $F:X\to I$ such that $f\left(y\right)=F\left(y\right)$ for all $y\in Y$. So $r$ is surjective.

We now want to show that $r$ is also continuous.

A subbase for $C(Y,I)$ consists of all $V_Y(K,U)\subset C(Y,I)$ where $K\subset Y$ is compact in $Y$ and $U\subset I$ is open, and $f\in V_Y(K,U)$ iff $f\left(K\right)\subset U$.

Note that since $Y$ has the subspace topology in $X$, these $K$ must be compact in $X$ as well. So define $V_X(K,U)$, consisting of continuous functions $F$ from $X$ to $I$ with $F\left(K\right)\subset U$. Then $p^{-1}\left(V_Y\right(K,U\left)\right)=V_X(K,U)$, 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 $r$ itself is continuous.

Compact subsets of $C(Y,I)$ have empty interior

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

The subbasis for the topology on $Y$ consists of sets of the form $V(K,U)$, where $K$ is compact in $Y$, $U$ is open in $I$, and $f\in V(K,U)$ iff $f\left(K\right)\subset U$.

The compact subsets of $Y$ are a) finite collections of points $y_i$, $i<\infty$, b) collections of points that include all points $y_i$ for $i>N$, along with $y_{\infty }$, and c) $y_{\infty }$ itself.

Let $W$ be an intersection of finitely many $V(K,U)$. If $W$ is non-empy, it is defined by an $N\in N$ and a collection of sets $\{U_i{\}}_{i<N}\cup \{U_{<\infty }\}\cup \{U_{\infty }\}$ open in $I$. Then $f$ belongs to $W$ iff:

• $\forall i<N:f\left(y_i\right)\in U_i$,

• $\forall i\ge N:f\left(y_i\right)\in U_{<\infty }$,

• $f\left(y_{\infty }\right)\in U_{\infty }$.

Since $W$ is non-empty and $f$ is continuous, we must have $\overline{U_{<\infty }}$ and $U_{\infty }$ having non-zero intersection. This actually implies that $U_{<\infty }$ and $U_{\infty }$ have non-zero intersection.

Note that this includes all possible $W$, as we can always set $U_i=I$ (or $U_{<\infty }=I$ or $U_{\infty }=I$), thus removing that particular restriction.

We want to show that such an $W$ cannot be equicontinuous at $y_{\infty }$.

To do so, fix points $u_i\in U_i$ for $i<N$, and $u_{\infty },u^{\prime }\in U_{<\infty }\cap U_{\infty }$, with $u_{\infty }\ne u^{\prime }$

Then consider the functions $f_m\in W$, for $m>N$, defined by:

• $f_m\left(y_i\right)=u_i$ for $i<N$,

• $f\left(y_i\right)=u^{\prime }$ for $N\le i\le m$,

• $f\left(y_i\right)=u_{\infty }$ for $i>m$.

These $f_m$ are all in $W$, and continuous (since $Y$ is discrete except at $y_{\infty }$). If $W$ were equicontinuous then for all $ϵ>0$, there would exists an open set $U$ in $Y$ containing $y_{\infty }$ such that $|f_m\left(y\right)-f_m\left(y_{\infty }\right)|<ϵ$ for all $m$ and all $y\in U$.

Since $U$ is open, there exists a $y_n\in U$, $n\ne \infty$. Set $ϵ=|u^{\prime }-u_{\infty }|/2$; then $|f_n\left(y_n\right)-f_n\left(y_{\infty }\right)|=|u^{\prime }-u_{\infty }|>ϵ$.

Since $W$ is not equicontinuous, and $Y$ is both compact and Hausdorff (and hence pre-compact Hausdorff), $W$ cannot be contained in a compact subset of $C(Y,I)$, by Ascoli’s Theorem.

Any open set in $C(Y,I)$ consists of unions of sets of the type $W$, so no (non-empty) open set is contained in compact subset of $C(Y,I)$.

$C(Y,I)$ is a Baire space

By the Baire category theorem, if $C(Y,I)$ is a complete metric space, then it is a Baire space.

Since $Y$ is compact, the compact-open and the uniform convergence topology match up on $C(Y,I)$. So $C(Y,I)$ has a uniform norm $d$ compatible with its topology: $d(f,g)={sup}_{y\in Y}|f\left(y\right)-g\left(y\right)|$.

So $C(Y,I)$ is thus a metric space. Since $I$ is complete, the uniform limit theorem shows $C(Y,I)$ is a complete metric space (and hence a Baire space).

There is no surjective continuous map $s:X\to C(X,I)$

Let $s:X\to C(X,I)$ be any continuous map.

Since $X$ is $\sigma$-compact, $X={\bigcup }_{i\in N}{K}_i$ for compact $U_i$. The set $r\circ s\left(K_i\right)$ is compact in $C(Y,I)$, since the continuous image of compact spaces are compact.

Let $L_i=C(Y,I)-r\circ s\left(K_i\right)$. Since compact sets in $C(Y,I)$ must have empty interior, the complement of $L_i$ has empty interior. Therefore the $L_i$ are dense in $C(Y,I)$.

The space $r\circ s\left(X\right)$ is equal to:

• ${\cup }_{i\in N}(r\circ s(K_i\left)\right)={\cup }_{i\in N}(L_i{)}^c={\left({\cap }_{i\in N}{L}_i\right)}^c$, where $c$ denotes taking the complement in $C(Y,I)$.

But, since $C(Y,I)$ is a Baire space, ${\cap }_{i\in N}{L}_i$ is dense, hence certainly not zero, and so $r\circ s\left(X\right)\ne C(Y,I)$.

Since $r$ is surjective $C(X,I)\to C(Y,I)$, the $s$ cannot be surjective onto $C(X,I)$, proving the result.

Removing the $T_4$ condition

We only used $T_4$ to prove the properties of the subsequence $Y$. Without $T_4$, we can use the continuous surjection $s:X\to C(X,I)$ itself. So now assume that such an $s$ exists, and we shall try and find a contradiction.

The $\sigma$-compact property implies $X$ is Lindelöf. Lindelöf is preserved by continuous maps, so because of $s$, $C(X,I)$ is also Lindelöf.

Because $I$ is $T_3$, then so is $C(X,I)$. But any Lindelöf regular space is normal (see theorem 14), hence, since $C(X,I)$ is also $T_2$, then it must be Hausdorff normal, ie $T_4$.

Now $C(X,I)$ contains at least the constant functions, so must be of uncountable cardinality. Since $X={\bigcup }_{i\in N}{K}_i$ is a countable union of spaces, there exists a $K_i$ such that $s\left(K_i\right)$ contains a set $W$ with infinitely many points. Pull these infinitely many points back to $K_i$, using the axiom of choice to choose a set $V\subset K_i$ such that for each $w\in W$, there is a unique $v\in V$ with $s\left(v\right)=w$.

Now, $K_i$ is compact, which means that it is countably compact. Now, Fréchet–Urysohn is hereditary, meaning that it applies to subspaces as well, so $K_i$ 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 $V$, and, passing to a subsequence if necessary, we can find a sequence $\{y_i{\}}_{i\in N}$ converging to a point $y_{\infty }$ that is not equal to any of the $y_i$. Define $Y$ as before.

The set $s\left(Y\right)$ must also be a convergent subsequence.

Let $f\in C(Y,I)$, then $f\circ s^{-1}\in C\left(s\right(Y),I)$ (note that $s^{-1}$ is well defined here, as it is a bijection between $Y$ and its image). Because $C(X,I)$ is itself $T_4$, there exists a function $F\in C\left(C\right(X,I),I)$ that is equal to $f\circ s^{-1}$ on $s\left(Y\right)$. Then the function $F\circ s$ is equal to $f$ on $Y$, and is an element of $C(X,I)$. Thus the restriction map $r:C(X,I)\to C(Y,I)$ is still surjective, and the topology on $Y$ (indeed on any sequence tending to a point) is the same as before. The argument for $r$ being continuous goes through as before.

Then the rest of the proof proceeds as above.