Manifolds: topological preliminaries

2020-01-20

Manifolds are mathematical spaces with two layers of structure imposed on them: a topological structure and a smooth structure. This post is about the topological structure, and will review the topological properties of manifolds and explore some of their consequences. To say that a manifold $M$ has topological structure means that it has a topology $\tau_M$ associated with it. To review, a topology of a set $M$ is a collection of subsets of $M$: $$\tau_M \subseteq \mathcal{P}(M)$$ The elements of $\tau_M$ are called the open sets, and they have the following three properties:

arbitrary unions of open sets are open: $\cup_\alpha \tau_\alpha \in \tau$
finite intersections of open sets are open: $\cap_{i=1}^N \tau_i \in \tau$
$\varnothing, M \in \tau$

A topological manifold of dimension $n$, $(M, \tau_M)$, is a special type of topological space with three defining topological properties:

it is a Hausdorff space
it is second-countable
it is locally Euclidean of dimension $n$

Hausdorff Spaces

A topological space $(M, \tau_M)$ is Hausdorff if for each two distinct points $p, q \in M$ there exist disjoint open sets $U, V$ such that $p \in U$ and $q \in V$. This immediately implies a very useful property: $$ \boxed{ \text{convergent sequences have unique limits}} $$ To see why this is true, recall that for general topological spaces a sequence $s_n$ converges to a point $s$ if for each open set $U$ containing $s$ there exists a number $N$ such that $s_n \in U$ for $n > N$. Our sequence cannot also converge to another point $t$ because from the Hausdorff property we know that there exists some open set $V$ which is disjoint from one of the open sets $U$ containing $s$. Because $U$ and $V$ are disjoint, and $s_n \in U$ for all $n$ greater than some $N$, $s_n \notin V$ for $n$ greater than $N$, and therefore $s_n$ cannot converge to $t$.

Another important property of Hausdorff spaces is: $$ \boxed{ \text{ compact sets are closed}} $$ This feels familiar to the Heine-Borel theorem, which says that compact sets in $\mathbb{R}^n$ are closed and bounded. (the actual theorem goes both ways). Hausdorff spaces capture part of that magic. A couple other properties of Hausdorff spaces are:

point sets are closed (sets containing a single point)
if $X$ is Hausdorff and $A \subseteq X$, then $A$ with the subspace topology is Hausdorff.
if $X$ and $Y$ are Hausdorff, then $X \times Y$ is Hausdorff with the product topology.

Note also that metric spaces are Hausdorff. To go the other way (and deduce that a space is metrizable), you need some other properties. More on that later.

Second Countability

A topological space is second countable if it has a countable base (also called a basis). A basis $\mathscr{B}$ of a topology $\tau_M$ is a collection of subsets of $M$ such that every open set $U \in \tau_M$ can be expressed as a union of basis sets: $\cup_\alpha B_\alpha$ where each $B_\alpha \in \mathscr{B}$. Having a basis for a topology is often useful. For instance, if you want to show that a function $f: X \rightarrow Y$ is continuous, you only need to show that $f^{-1}(B)$ is open in $X$ for each $B \in \mathscr{B}$, the basis of $Y$. To see why this works, write: $$ f^{-1} \left( U \right) = f^{-1} \left( \cup_\alpha B_\alpha \right) = \cup_\alpha f^{-1} \left( B_\alpha \right)$$ And since the union of open sets is open, we can conclude that $f^{-1} \left( U \right)$ is open if each $f^{-1}(B_\alpha)$ is open, and therefore that the function is continuous.

An example of a basis for $\mathbb{R}^n$ is the set of all open balls of positive real radius centered at every point in $\mathbb{R}^n$. This basis has a cardinality $\left| \mathbb{R}^n \times \mathbb{R} \right| = \left|\mathbb{R}\right|$. However, there exists a basis for $\mathbb{R}^n$ which has only countably many elements: the set of all open balls of rational radius centered at points in $\mathbb{Q}^n$. This has cardinality $\left| \mathbb{Q}^n \times \mathbb{Q} \right| = \left| \mathbb{Q} \right|$ and hence is countable. Therefore $\mathbb{R}^n$ is second countable, since it has a countable base. Topological spaces with this property have a number of nice properties. For instance, if $M$ is second countable: $$ \boxed{M \text{ has a countable dense subset}} $$ One can construct such a countable dense subset $D$ by simply selecting one element from each basis set: $$ D = \{ \text{select one } x_n \in B_n \ : \ B_n \in \mathscr{B} \} $$ This guarantees that every open set $U \in \tau_M$ contains an element of $D$, and so every point in $M$ is an adherent point of $D$. The closure of $D$ is therefore $M$, and $D$ is countable because $\mathscr{B}$ is countable.

Locally Euclidean of dimension $n$

This means that every point of $M$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^n$. This is maybe the most important property of topological manifolds. The existence of these homeomorphisms allows you to import lots of topological structure from $\mathbb{R}^n$ into $M$. Recall that a homeomorphism between two topological spaces $X$ and $Y$ is a bijection $\varphi : X \rightarrow Y$ which is continuous, and for which $\varphi^{-1}: Y \rightarrow X$ is continuous. From the definition of continuity, one can see that such a function gives a bijection between the open sets of $X$ and $Y$. Since topological properties are defined in terms of the open sets, this means that the two homeomorphic spaces $X$ and $Y$ will share topological properties. If two spaces are homeomorphic, they are:

Both compact or neither compact
Both Hausdorff or neither Hausdorff
Both connected or neither connected

We can now provide a rough sketch for why topological manifolds are metrizable (there exists a metric which induces their topology). Urysohn's metrization theorem states that Hausdorff, second-countable, regular spaces are metrizable (recall that a regular space is one for which any closed set $C$ and a point $p$ not in $C$ admit disjoint open neighborhoods). It turns out that spaces which are Hausdorff and locally compact are regular. Recall that locally compact means that every point in the space has a compact neighborhood. Consider some point $p \in M$. Since $M$ is locally Euclidean of dimension $n$, there exists some neighborhood $U$ of $p$ homeomorphic to an open set $\hat{U} \in \mathbb{R}^n$. Consider the point $\varphi(p) \in \mathbb{R}^n$. Since $\hat{U}$ is open, there is some open ball $B_\epsilon(\varphi(p))$ contained in $\hat{U}$, and inside of this we can find a closed ball $\overline{B}_{\epsilon/2}(\varphi(p))$, which is compact by the Heine-Borel theorem. Then: $$ \varphi^{-1}\left( \overline{B}_{\epsilon/2}(\varphi(p))\right) $$ is a compact set in $U$ containing $p$. We have essentially used the fact that $\mathbb{R}^n$ is locally compact to conclude that $M$ is locally compact. Since $M$ is locally compact, and therefore regular, by Urysohn's metrization theorem, all topological manifolds are metrizable!