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.

Follow

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.

latex

@Vierkantor this is really clever! I ended up solving it differently, by pawning off a lot of the work on a theorem proven for me in text, but...

let $M$ be a topological n-manifold, $\mathcal{B}$ be the collection of all coordinate balls on $M$

we know b/c $M$ is locally Euclidean that for every point $x \in M$, there exists an open neighborhood $U$ homeomorphic to an open subset of $\mathbb{R}^n$, by the homeomorphism $\phi$. so fix $x$ and $U$ as described.

then, let $B \subseteq \phi(U)$ be an open ball in $\mathbb{R}^n$ contained within the image of $U$, where $\phi(x) \in B$. (we know this exists because $\mathbb{R}^n$ has a basis of balls.

then, $\phi^{-1}(B)$ is a coordinate ball, where $x \in \phi^{-1}(B)$.

so because $x$ was an arbitrary point in $M$, we know that there is an element of $\mathcal{B}$ such that $x$ is in one of its elements for each $x \in M$. so $\mathcal{B}$ is a cover!

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 $\mathcal{B}$ covers every n-manifold.

as you say, to show an open $U$ is a union of elements in $\mathcal{B}$, it suffices to show that each $x \in U$ has an open neighborhood $x \in N \subseteq U$, such that $N \in \mathcal{B}$.

so if we fix an open subset $U \subseteq M$, then WLOG we can state that since $U$ is an n-manifold, the subset of $\mathcal{B}$ with elements contained in $U$ is a cover for $U$. so because each element of this subset is an open neighborhood for every point in $U$, the above criterion is satisfied, and we have a basis.

qed?

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! 💡

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

@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.comhint 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.