Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I hope I'm using the terminology correctly. What I mean is this: fix $K = \mathbb{R}$ or $\mathbb{C}$ (I'm interested in both cases). Which topological spaces $X$ have the property that for every open set $U$, every continuous function $f : U \to K$ is a quotient of continuous functions $\frac{g}{h}$ where $g, h : X \to K$ and $h \neq 0$ on $U$?

share|improve this question
1  
Don't you mean that $h$ is nowhere vanishing on U? –  Yemon Choi Nov 20 '09 at 6:40
    
As topological spaces $\mathbb C = \mathbb R \times \mathbb R$, and so for the purposes of detecting spaces there is no difference. –  Theo Johnson-Freyd Nov 20 '09 at 8:03
2  
@Theo: we are of course using the multiplicative structure on K, at least in the question. So while there might turn out to be no difference I'm not convinced it's for the reason you give. –  Yemon Choi Nov 20 '09 at 8:07
1  
@Theo: while it's always nice to see love for Gelfand-Naimark, I don't quite understand its relevance to Qiaochu's question... (Also, C*-algebras are very rigid and non-sheafy objects) –  Yemon Choi Nov 20 '09 at 8:21
2  
For locally compact hausdorff spaces this is equivalent to finding h continuous on X such that fh extends to a continuous function up to the boundary, since than you can extend the product to a g continuous on the full space. –  Gian Maria Dall'Ara Nov 20 '09 at 11:27
show 7 more comments

2 Answers 2

up vote 5 down vote accepted

This isn't a complete answer, but I think that whatever the family is, it contains compact metric (metrisable) spaces. With a paracompactness argument, I suspect that it would extend to locally compact, and I would not be surprised if one could replace "metrisable" by something weaker (though I think that it would need that separation property one-above-normal which I can never remember the name of: namely that every closed set is the zero set of a continuous function).

Here's a proof (I hope): Let $M$ be a compact metric space, $U \subseteq M$ an open subset, $f : U \to \mathbb{R}$ a continuous function. Let's write $K$ for the complement of $U$ in $M$. For each $n \in \mathbb{N}$, let $C_n \subseteq U$ be the subset consisting of points at least distance $1/n$ away from $K$. Then $C_n$ is closed in $M$, hence compact, and $\bigcup C_n = U$. Let $h_0 : M \to \mathbb{R}$ be the "distance from $K$" function (so that $C_n = h_0^{-1}([1/n,\infty))$). Let $V_n$ be the complement of $C_n$.

As $C_n$ is compact, $f$ is bounded on $C_n$. Let $a_n = \max\{|f(x)| : x \in C_n\}$, then $(a_n)$ is an increasing sequence. Let $(b_n)$ be a decreasing sequence that goes to $0$ faster than $(a_n)$ increases, specifically that $(a_nb_n) \to 0$. Let $r : [0,\infty) \to [0,\infty)$ be a continuous decreasing function such that $r(1/n) = b_{n+1}$ (as $(b_n) \to 0$ (this always exists) and let $h = r \circ h_0$. Then for $x \in V_{n-1}$, $h_0(x) \lt 1/(n-1)$ so $h(x) \lt b_n$.

Then $h : M \to \mathbb{R}$ is a continuous function. Moreover, $h f$ (the product, with $h$ restricted to $U$) has the property that for $x \in C_n \setminus C_{n-1} = V_{n-1} \setminus V_n$,

$$ |(f h)(x)| = |f(x)| |h(x)| \le a_n b_n $$

Thus as $x \to K$, $(f h)(x) \to 0$ and so we can extend $f h$ to a continuous function $g : M \to \mathbb{R}$ by defining it to be $0$ on $K$.

Then on $U$, $f = g/h$.

(I made this up, so obviously, there may be something I've overlooked in this so please tell me if I'm not correct.)

Edit: This one's been bugging me all weekend. I've even gone so far as to look up perfectly normal.

This property holds for perfectly normal spaces. In a perfectly normal space, every closed set is the zero set of a function (to $\mathbb{R}$, and this characterises perfectly normal spaces according to Wikipedia).

Here's the proof. Let $X$ be a perfectly normal space. Let $U \subseteq X$ be an open set, and $f : U \to \mathbb{R}$ a continuous function. Let $r : X \to \mathbb{R}$ be such that the zero set of $r$ is the complement of $U$. Let $s : \mathbb{R} \to \mathbb{R}$ be the function $s(t) = \min\lbrace 1, |t|^{-1}\rbrace$.

The crucial fact is that if $p : U \to \mathbb{R}$ is a bounded function then the pointwise product $r \cdot p : U \to \mathbb{R}$ (technically, $p$ should be restricted to $U$ here) extends to a continuous function on $X$ by defining it to be zero on $X \setminus U$.

From this, the rest follows easily.

  1. The composition $s \circ f$ is bounded on $U$, hence $r \cdot (s \circ f)$ extends to a continuous function on $X$, say $h$.

  2. The product $(s \circ f) \cdot f$ is also bounded on $U$, since $(s \circ f)(x) = \min\lbrace 1, |f(x)|\rbrace)$. Hence $r \cdot (s \circ f) \cdot f$ extends to a continuous function on $X$, say $g$.

  3. As $s(t) \ne 0$ for all $t \in \mathbb{R}$, $(s \circ f)(x) \ne 0$ for all $x \in X$. Hence $h(x) \ne 0$ for all $x \in U$.

  4. Finally, on $U$, $g(x) = h(x) \cdot f(x)$, whence, as $h$ is never zero on $U$, $f = g/h$ as required.

This isn't a complete characterisation of these spaces. Essentially, this result holds if there are enough continuous functions (as above) on $X$ and if there are too few.

As an example of the latter, consider a topological space $X$ where every pair of non-trivial open sets has non-empty intersection. Then there can be no non-constant functions to $\mathbb{R}$, either on $X$ or on any open subset thereof. Hence every continuous function on an open subset of $X$ trivially extends to the whole of $X$.

However, there's probably some argument that says that once you have sufficient continuous functions (say, if the space is functionally Hausdorff - i.e. continuous functions to $\mathbb{R}$ separate points) then it would have to be perfectly normal. The difficulty I have with making this into a proof is that there's no requirement that the function $g$ be zero on the complement.

Finally, note that metric spaces are perfectly normal so this supersedes my earlier proof. I leave it up, though, in case it's of use to anyone to see the workings as well as the current state. (Actually, for the record I ought to declare that initially I thought that this was false for almost all spaces. However, once I'd examined my counterexample closely, I realised my error and now I'm having difficulty thinking of a reasonable space where it does not hold.)

share|improve this answer
    
Metrizability seems too much. After all, what you use is sigma compactness of every open set and the existence of the function going to zero fast enough. If K_n is the exhaustion of U by compact sets than a positive map whose value is b_n on Cl(K_n\K_n-1) and 0 outside Int(K_n+1\K_n-2) should be enough. So I think the right condition, at least for your argument to work is the existence of bump functions (loc comp hausdorff) and sigma compactness of open sets. –  Gian Maria Dall'Ara Nov 20 '09 at 13:30
    
Sigma compactness is a natural condition, which is useful even in measure theory since it ensures the regularity of measures given by Riesz representation theorem. –  Gian Maria Dall'Ara Nov 20 '09 at 13:30
    
I suspect that sigma compactness is too much as well, or at least with paracompactness can be replaced by a sort of local sigma compactness. I'll be interested to learn if there's a simple description of all the topological spaces that have this property. However, for this answer I just wanted something that would work and wasn't too restrictive. –  Loop Space Nov 20 '09 at 13:51
    
Note that I got rid of both metrisability and sigma compactness now. –  Loop Space Nov 22 '09 at 21:13
    
Very nice proof! Now, even if it is not a characterisation, it's much more satisfying. –  Gian Maria Dall'Ara Nov 22 '09 at 21:49
show 2 more comments

I tried a bit of thinking, but I haven't worked all the details. I have a hint though that may lead to the answer of your question. You may want to regard the continuous functions over an open set as a ring. This ring is reduced and commutative (thus there is a so-called rational completion) and we could then look at rational completion of them and this may lead to an answer.

A good and downloadable reference of this is found here. A classical reference (and also the best one) is the book of Lambek "Lectures on Rings and Modules" by Lambek (please don't confuse it with the book of Lam, who happens to have the same first 3 letters in his last name, entitled "Lectures on Modules and Rings"), see for instance sections 2.3 and 4.4 of the book.

A few years ago, I had written a small entry in Planetmath that characterized rational extensions of commutative reduced rings. And you can use that as an easy definition.

share|improve this answer
1  
Note: I know the OP asks about the topology of X. But to study the the rational extensions of continuous function, information about Spec C(U) is passed (C(U) is the ring of cont's functions from U to K) and there is a relation between the topology of Spec C(U) and the topology of U. –  Jose Capco Nov 21 '09 at 1:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.