# Can we get un-obvious results by defining sophisticated topologies?

What I originally found so interesting about general topology was how general a type of thing a topology is, and how the terminology open, closed, compact, continuous, convergence et cetera means something completely different depending on which topology we are using. It seemed to me that each result was actually one result PER choice of topology. But the course I did mostly focused on real space and the Euclidean topology, the boundedness principle and equivalence of norms. I admit all of these are practically useful, but they were not what made topology seem so interesting.

My question is this: Are there any cases where we engineer an unusual topology with an eye to proving specific unobvious results. The work would then be in proving it is a topology, exactly what continuous functions look like, what compactness is and so on. And then invoke a general theorem to get the result?

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The Ellentuck topology has some connection with Ramsey Theory. – William Sep 4 '13 at 12:53
Search: Krull topology, Galois theory – Amr Sep 4 '13 at 12:59
See the answers to this question: math.stackexchange.com/questions/456309/… – Amr Sep 4 '13 at 13:02
The Zariski topology and related topologies come up in algebraic geometry. – Gyu Eun Lee Sep 4 '13 at 13:06
In the last chapter of Oxtoby's Measure and Category he defines a topology on the reals such that sets of measure zero and nowhere dense sets are the same! – Mauricio G Tec Sep 4 '13 at 19:17

Let $G$ be an infinite graph, such that every finite subgraph of $G$ can be colored with $k$ colors. Is it necessarily the case that $G$ can be colored with $k$ colors?

The answer is yes, and to prove this we will begin by defining a topology on the space of all colorings of $G$.

Let $V$ be the set of vertices of $G$. A coloring of $G$ - not necessarily legal - is just a function $f : V \to A$, where $A$ is the finite set of $k$ colors. Let us denote the space of all colorings by $X$.

We can topologize $X$ by taking a basic open set to be of the form $U_{v,c}$ where $v$ is a vertex of $G$, $c$ is some color in $A$, and $U_{v,c}$ is the set of those colorings $f \in X$ such that $f(v)=c$. A little thought reveals that this is just the space product space $A^V$ where $A$ is given the discrete topology.

There is a highly nontrivial theorem of Tychonoff that says that the product of compact spaces is again compact. So the space $X$ is compact. This implies that every collection of closed subsets of $X$ with the finite intersection property (the intersection of finitely many sets in the collection is nonempty) has a nonempty intersection.

Let $H$ be a finite subgraph of $G$. If $f$ is a coloring of $G$, we can obtain a coloring of $H$ by restricting $f$ to the set of vertices of $H$. Of course, if $f$ were a valid coloring of $G$, then restricting it to $H$ gives a valid coloring of $H$ - in $H$, there are less vertices and edges to worry about.

Conversely, let $f$ be a coloring of $G$, and suppose that restricting $f$ to each finite subgraph of $G$ gives a valid coloring. I claim that $f$ is a valid coloring of $G$, too. Suppose not. Then there are vertices $u,v$ connected by an edge such that $f(u) = f(v)$. But then the restriction of $f$ to the finite subgraph $H$, which consists of the vertices $u,v$ and the edge between them, is not a legal coloring. This contradicts our initial asumption.

So we are going to find a coloring of $f$ of $G$ whose restriction to each finite subgraph is legal and this will prove our theorem.

If $H$ is a finite subgraph of $G$, let $C_H$ denote the set of those colorings of $G$ whose restriction to $H$ is legal. By hypothesis $C_H$ is nonempty for every such $H$. Furthermore, the set $C_H$ is closed. For if $f$ is a coloring of $G$ which lies outside $C_H$, then there are vertices $u,v$ of $H$ connected by an edge in $H$ such that $f(u) = f(v)$. The set of $U$ of those $g$ which agree with $f$ on $u,v$ is open and it contains $f$, but it is disjoint from $C_H$ (no such $g$ can be valid when restricted to $H$). It follows that $C_H$ is closed, since its complement contains a neighborhood of each of its points.

Finally, the family ${C_H}$ enjoys the finite intersection property: if $H_1, ..., H_n$ are finite subgraphs, then a valid coloring of their union is a valid coloring of each of them individually. So $C_{H_1 \cup ... \cup H_n}$ is a nonempty subset of $C_{H_1} \cap ... C_{H_n}$.

By compactness the intersection of all the $C_H$, for $H$ a finite subgraph, is nonempty. If $f$ is an element of the intersection, then it is a valid coloring of each finite subgraph by definition, and hence, by the above, a valid coloring of $G$.

The above proof is very similar to the proof that the compactness theorem in topology yields the compactness theorem for propositional calculus; thus logical and topological compactness are distinct, but related notions.

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I see that $C_H$ has the finite intersection property, but not why this follows from the previous line. I also don't see how we conclude the final result, that there exists a legal colouring for the infinite graph. – Daron Sep 4 '13 at 14:48
I edited accordingly. The most subtle point I think is why these sets are closed. By the way - a very similar argument proves the compactness theorem for propositional calculus. – Yuval Sep 4 '13 at 14:59
I do not see why your addition gets us any closer to concluding that the infinite intersection $\cap_{|H|< \infty, H \subset G} C_H$. Maybe I am looking at this the wrong way, could you please explain further? – Daron Sep 4 '13 at 15:37
I hope its clear now – Yuval Sep 4 '13 at 16:10
It is, thanks. The only thing that was really missing is " This implies that every collection of closed subsets of X with the finite intersection property (the intersection of finitely many sets in the collection is nonempty) has a nonempty intersection." I can see why that is true. – Daron Sep 4 '13 at 21:07

What you've written sounds like a very good description of the entire field of functional analysis. One constructs topologies on various spaces of functions and uses topological properties to prove things.

For instance, if one is looking for a function with certain properties (say, one that solves a certain differential equation), a common technique is to construct a sequence of functions that "approximately" have those properties, and use compactness to claim the sequence has a limit point. Then, using some sort of continuity, claim that the limit point has exactly the desired properties. Often the topology has been specifically constructed to provide such continuity.

Examples include $L^p$ spaces, Sobolev spaces, spaces of continuous functions and of distributions (generalized functions), and many more.

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Glad to know this. This makes me interested in self-studying functional analysis in the near future +1 – Amr Sep 4 '13 at 23:11

Here's a great example of what you're looking for: Furstenberg's proof of the infinitude of primes. http://en.wikipedia.org/wiki/Furstenberg%27s_proof_of_the_infinitude_of_primes

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I have seen this proof before, and I was considering mentioning it in the question. It is not exactly what I am looking for, since while it is written using topological language, it could does not make use of any general results, and could equally be re-expressed so as to be understandable to someone not familiar with topology. Though I admit familiarity with topological axioms does help in reading it. – Daron Sep 4 '13 at 12:58

André Weil imagined what would happen if a nonsingular projective algebraic variety over a finite field had a topology which satisfied certain properties (i.e. had a "good" cohomology theory, whatever that is). Using such an assumption, he was able to prove extremely important results about counting the number of points on such varieties (over a finite field, the number of points is always finite). These were first conjectured in the special case of curves by Artin. The problem was that such a topology with a good cohomology theory was not known. The Weil conjectures were born.

Weil was able to prove the conjectures for curves but there was really no hope for general varieties of dimension $> 1$. Grothendieck defined a new topology largely for the goal of proving the Weil conjectures. This is not exactly a topology, it is what is known as a Grothendieck topology. The gist of it is that the most important thing (in algebraic geometry) is that a topology allows you to speak of covers and refinements of covers. By only requiring a "topology" to satisfy axioms relating to covers, one is allowed more flexibility. With this flexibility in mind, Grothendieck defined and developed several topologies and cohomologies, most notably the étale topology and I-adic topology.

By mostly proving theorems for these topologies which parallel well known theorems for more "usual" topologies, a twenty or so year program was able to prove most of the Weil conjectures. Only the Riemann hypothesis remained. Grothendieck outlined a program of standard conjectures which would ultimately lead to a proof of the Riemann hypothesis. These remain largely open. Deligne was able to prove the Riemann hypothesis using a completely different argument which did not parallel well-known theorems in "usual" topology. He was awarded the fields medal for his proof.

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Topologies in the sense of Grothendieck are not at all like topologies in the sense of general topology... – Zhen Lin Sep 4 '13 at 15:36
This is specifically mentioned in my answer. I did consider this as perhaps meaning it does not answer the question, but this is (IMO) THE example of an "unusual topology" proving extremely deep results. I believe this trumps the fact that it is not a really a topology. – RghtHndSd Sep 4 '13 at 15:40
While it may have been inspired by topologies and therefore have some points in common with them, it is not a topology; this trumps everything else, as far as I’m concerned. – Brian M. Scott Sep 5 '13 at 5:23
@BrianM.Scott: I don't find this an unreasonable position, it just isn't mine. Please, downvote if you agree with Brian and I'll remove if it goes into the negative. – RghtHndSd Sep 5 '13 at 13:02