@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.)
@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 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.
@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?
@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 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.
@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.
@Vierkantor it does, because coordinate balls are coordinate domains and coordinate domains are open subsets. but yeah, alright
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