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Merely In The Name Of Self-Outward Violence

@bubbloquacious / bubbloquacious.tumblr.com

Roos ∗ They/Them ∗ Seventh Year Math Student ∗ Main is @bubbliterally but I talk more here ∗ An outlet for my mathematical ramblings >:^) ∗ Icon and banner are by @tamberella

The Topology Game

Here's an updated list of topological properties that I use to play @topoillogical's Topology Game. The rules are as follows. You roll a random integer between -N and N inclusive, where N is the number of properties below (currently N = 114). If the number is negative, you get the negation of the property whose number is the magnitude of the number. You come up with a topological space satisfying this property. This was round 1.

For all future rounds, you start by rolling another property. If the new property follows from the previous properties, prove this. If the new property contradicts the previous properties, prove this. If the new property is independent of all the previous, come up with two spaces satisfying all the previously rolled properties, where one space has the new property and one space has its negation.

You 'win' the game if you manage to solve round 10. You can never truly lose as you can always come back to a given game.

List of 114 properties:

Manifolds are so damn complicated. Some topological manifolds have no smooth structure at all. If a manifold has a k-time differentiable smooth structure for k ≥ 1, then it has an n-time differentiable smooth structure for every n. If the manifold is second-countable and Hausdorff, then these smooth structures are unique, but if it fails either of them then for any k and n there exist multiple non-diffeomorphic structures.

The complex numbers and the split-complex numbers both have only two automorphisms that keep the reals fixed. But scaling the nilpotent axis in the dual numbers by any non-zero real scalar is always an automorphism. Weird!

So if your coordinates lie in an integral domain, the union of the solutions sets of two algebraic equations is itself the solution set of an algebraic equation, because xy = 0 if and only if x = 0 or y = 0. In this setting an algebraic set is a set that's the simultaneous solution set of a system of algebraic equations, which you can show pretty easily to be the same thing as the vanishing set of an ideal of polynomials. Algebraic sets are then closed under arbitrary intersections and finite unions, so they're the closed sets in a topology which we call the Zariski topology.

In other algebraic settings, though, this doesn't hold up! If you take two solutions sets of linear equations in some vector space, two lines through the origin for example, then their union is not generally the solution set of a linear equation, because we know that such sets are linear subspaces.

C'mon guys! This is important* stuff!

*zing! got you again with another one of my trademark 'unnecessary generalizations'

So if your coordinates lie in an integral domain, the union of the solutions sets of two algebraic equations is itself the solution set of an algebraic equation, because xy = 0 if and only if x = 0 or y = 0. In this setting an algebraic set is a set that's the simultaneous solution set of a system of algebraic equations, which you can show pretty easily to be the same thing as the vanishing set of an ideal of polynomials. Algebraic sets are then closed under arbitrary intersections and finite unions, so they're the closed sets in a topology which we call the Zariski topology.

In other algebraic settings, though, this doesn't hold up! If you take two solutions sets of linear equations in some vector space, two lines through the origin for example, then their union is not generally the solution set of a linear equation, because we know that such sets are linear subspaces.

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