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.

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 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 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 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.

re: latex

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

re: latex

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

re: latex

@Vierkantor nice ok cool it is time for me to attempt the Final Problem Of This Chapter

Anne `hex(44203)` Baanen@Vierkantor@mastodon.vierkantor.comre: 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.