What is the basis of a Manifold?

Question

I am taking a course on Manifolds, and we glossed over the definition of a topological manifold without much discussion. The thing I am specifically unsure about, is what the basis of a manifold is.

From previous intuition, I was thinking about a basis akin to that of a vector space. But we do not have global coordinates and so clearly this is the wrong interpretation.

Doing a bit of research, it seems that the basis is a topological basis. But I do not fully understand that in this context. Is our basis the collection of all points in our manifold (which seems unlikely because then it would rarely be countable?), or is it the collection of open sets?

If it is the collection of open sets, then will each open set be contained in a chart of the maximal atlas? Any elaboration on the nature of this basis would be very helpful in my understanding.

Answer 1

A base (basis is usually reserved for algebraic uses of the word) for $X$ is set $\mathcal{B}$ of open subsets of $X$ such that for every open set $O$ and every $x \in O$ we have some $B \in \mathcal{B}$ such that $x \in B \subseteq O$. It turns out that this notion is quite handy to discuss topological properties: continuity of a function can be determined by looking at base elements e.g. and in a manifold the locally Euclidean (homeomorphic to $\Bbb R^n$ for some $n$) open sets form a base by definition. In many cases restrictions are put on the size of a base, especially spaces with countable bases are a popular subclass of spaces: this implies that the space is Lindelöf, separable etc. and makes it more amenable to analysis techniques (we can use countable series, countable covers by locally Euclidean neighbourhoods etc.) So very often (in many texts) manifolds are assumed to have a countable base (as all $\Bbb R^n$ have). A maximal atlas is always a base and if the manifold has a countable base, we can reduce any atlas to some countable atlas (which makes it more manageable), by standard theorems in general topology.

Answer 2

The definition is referring to this kind of basis: https://en.wikipedia.org/wiki/Base_(topology)

To compare it with Linear Algebra, one property of a topological basis is that you can write any open set as a union of basis sets. The difference is that this union is not unique, like it is in Linear Algebra.

Also, one usually requires that Topological Manifolds have a countable basis, which is to make some technical properties work out nicely later on.

A concrete example of a countable basis for $\mathbb{R}$ is the collection of all open intervals of the form $(q - \varepsilon, q + \varepsilon)$ with $q \in \mathbb{Q}$ and $\varepsilon \in \mathbb{Q}_{>0}$.

Answer 3

A basis of a topological space is a set of open sets so that you get all open sets as unions of them. For example, in $\mathbb R^n$, the open balls form a basis of the standard topology.

The collection of all open sets obviously is a basis, but usually one wants a much smaller basis. For example, in $\mathbb R^n$, the open balls with rational center and radius form a countable basis (that is, it contains only countably many elements, but still gives all open sets as unions).

Answer 4

A topological space is determined by their "open sets", a vector space is determined by their "elements".

A basis of a topological space $X$ is a collection $\mathcal{B}$ of special subsets of $X$, just like a basis of a vector space $V$ is a collection $\mathcal{B}$ of special elements of $V$.

A basis of a vector space has the property that, given any "element $a\in V$" it can be written as "linear combination" of elements of the basis.

A basis of a topological space has the property that, given any "open set $U\subseteq X$" it can be "contained" in a "union" of elements of the basis.

The other property of basis of a vector space does not seem to have a nice relation with other property of basis of a topological space. Other property of a basis of a topological space is that, given two open sets $U_1$ and $U_2$ in the basis $\mathcal{B}$, for each $x\in U_1\cap U_2$, there exists a basis element $U_3$; that is an open set of $X$ contained in the collection $\mathcal{B}$, such that $x\in U_3\subseteq U_1\cap U_2$.