nhaliday + math.at   12

gn.general topology - Pair of curves joining opposite corners of a square must intersect---proof? - MathOverflow
In his 'Ordinary Differential Equations' (sec. 1.2) V.I. Arnold says "... every pair of curves in the square joining different pairs of opposite corners must intersect".

This is obvious geometrically but I was wondering how one could go about proving this rigorously. I have thought of a proof using Brouwer's Fixed Point Theorem which I describe below. I would greatly appreciate the group's comments on whether this proof is right and if a simpler proof is possible.

...

Since the full Jordan curve theorem is quite subtle, it might be worth pointing out that theorem in question reduces to the Jordan curve theorem for polygons, which is easier.

Suppose on the contrary that the curves A,BA,B joining opposite corners do not meet. Since A,BA,B are closed sets, their minimum distance apart is some ε>0ε>0. By compactness, each of A,BA,B can be partitioned into finitely many arcs, each of which lies in a disk of diameter <ε/3<ε/3. Then, by a homotopy inside each disk we can replace A,BA,B by polygonal paths A′,B′A′,B′ that join the opposite corners of the square and are still disjoint.

Also, we can replace A′,B′A′,B′ by simple polygonal paths A″,B″A″,B″ by omitting loops. Now we can close A″A″ to a polygon, and B″B″ goes from its "inside" to "outside" without meeting it, contrary to the Jordan curve theorem for polygons.

- John Stillwell
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october 2017 by nhaliday
Best Topology Olympiad ***EVER*** - Affine Mess - Quora
Most people take courses in topology, algebraic topology, knot theory, differential topology and what have you without once doing anything with a finite topological space. There may have been some quirky questions about such spaces early on in a point-set topology course, but most of us come out of these courses thinking that finite topological spaces are either discrete or only useful as an exotic counterexample to some standard separation property. The mere idea of calculating the fundamental group for a 4-point space seems ludicrous.

Only it’s not. This is a genuine question, not a joke, and I find it both hilarious and super educational. DO IT!!
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october 2017 by nhaliday
Covering space - Wikipedia
A covering space of X is a topological space C together with a continuous surjective map p: C -> X such that for every x ∈ X, there exists an open neighborhood U of x, such that p^−1(U) (the inverse image of U under p) is a union of disjoint open sets in C, each of which is mapped homeomorphically onto U by p.
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january 2017 by nhaliday

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