re: latex

@hazel ah ok, then I retract what I said and everything looks good!

re: latex

@hazel Your proof looks good to me! The only part I would be extra careful with is to ensure that the elements of B are *open* coordinate balls. I suspect that does follow from the definition of coordinate balls, but I'd have to check.

re: latex

@hazel
> now the weird part -- we know that any open subset of an n-manifold (imbued with the subspace topology) is itself an n-manifold.
so we know that covers every n-manifold.

ooh, that's a nice trick! 💡

hint 5 (QED)

@hazel But this last condition is exactly how the topology on R^n is defined, so let N' be an open ball around x contained in U', and set N = (φ_V)⁻¹(N'). Since φ_V is a bijection, we have x ∈ N \subseteq U \intersect V \subseteq U. And the restriction of φ_V to N → N' is again a homeomorphism, so N is an open (why?) neighbourhood of x contained in U that is a coordinate ball. Therefore B, the set of open coordinate balls, form a topology basis for X, QED.

hint 4

@hazel Let V be a neighbourhood of x which is homeomorphic to some open V' \subseteq R^n. Let's say the map is φ_V. Then U \intersect V is open in the subspace topology, so U' = φ(U \intersect V) is an open in V', and therefore also an open in R^n, containing x' = φ_v(x).

Since φ_v is a homeomorphism, it suffices to show that x' has a neighbourhood N' contained in U', such that N' is an open ball.

hint 3

@hazel So take as a basis B, the set of all open coordinate balls in X. We will show that for all opens U and x ∈ U, there is an open coordinate ball N that is contained within U.

What does this condition look like if we use the local homemorphism to (a subset of) R^n?

hint 2

@hazel To show an open U is a union of elements in B, it suffices to show each x ∈ U has a(n open) neighbourhood x ∈ N \subset U, such that N ∈ B.

hint 1

@hazel Note that a basis can be as big as we want it to be: if B is a basis and S is a set of open sets, then B \union S is still a basis.

So if we want to show a basis exists such that each open in the basis satisfies some property, then just take the set of all opens that satisfy the property, and show it forms a basis.

@hazel Been a while since I did topology, but I think I got it. Here's the definitions I'll use: a basis B is a subset of the set of all opens T, such that each U ∈ T is a union of elements in B. A topological manifold X (of dimension n) is a Hausdorff space such that each x ∈ X has a neighbourhood U which is homeomorphic to some U \subseteq R^n. (I believe these are standard defs.)

This random rooster kept showing up in my yard today, so I set up the most Wile E Coyote-ass trap to get rid of him

@cadadr I still struggle in English due to the lack of a morpheme which means "Sapir-Whorff is nonsense".

@trickster apart from the part where they didn't pay the workers :/

in conclusion, fyre festival was a land of contrasts.

garbage collector seems to work fine now, only we still run out of memory! apparently it freed 2.3 GB during execution before it couldn't keep up any more (the virtual machine has 1GB available)

some more bugfixing later, looks like we can run for a while until we get out of memory. in other words: time to re-activate the GC and see what happens! 