# Topological Manifolds: Definition and Properties

In this post, I would like to explore some basic properties of Topological manifolds.

As a pre-concept of Differential manifolds, the definition and properties of topological manifolds is quite important. In fact, we will emphasis on what kind of impact the locally Euclidean will make on the original topological space.
1. The Definition of Topological Manifolds Let’s first give the definition of Topological Manifolds.

Definition 1. Suppose that $M$ is a topological space. If
1. $M$ is Haussdorff. I.e., for any points $p,q$, $p\neq q$, then there are neighbourhoods $U, V$ of $p$ and $q$ respectively, such that $U\cap V=\emptyset$.
2. For each $x\in M$, there is a neighbourhood $U$ of $x$ and an integer $n\geq0$, such that $U$ is homeomorphic to $\R^n$.
Then, we call that $M$ is a Topological Manifolds.

Remark 1. condition 2 also called “Locally Euclidean”.

Remark 2. In condition 2, the neighborhood can be replaced by open neiborhood of $M$.

Proof . Since every open neighborhood is also an neibourhood, thus, if we assumed that the existence of open neighborhood $U$ for every $x$, then we also have the existence of the neighborhood of $x$.

On the other hand, suppose that for each $x\in M$, there exists a neighborhood $U$ of $x$, then, by the definition of neighborhood, we must have an open set $V$, such that $x\in V\subset U$. Now, we need the easy verified facts: If $\phi$ is a homeomorphic between the topological space $M$ and $N$, and $U$ is a subset of $M$, then under the induced topology, $\phi$ is a homeomorphic between $U$ and $\phi(U)\subset N$. Apply this fact, Since $\phi$ is a homemorphic between $U$ and $\R^n$, and $V\subset U$ is open, we know that $\phi(V)$ is also open in $\R^n$. Noted that $\phi(x)\in\phi(V)$, we can find an open ball $W$, $\phi(x)\in W\subset \phi(V)$, use again that $\phi^{-1}$ is a homeomorphic between $\R^n$ and $U$, we know that $\phi^{-1}(W)$ is open in $U$. It is apparently that $\phi^{-1}(W)\subset V\subset U$, and $\phi$ is a homeomorphic between $\phi^{-1}(W)$ and $W$. Now by the well known fact that any open ball in $\R^n$ is homeomorphic to $\R^n$, from which we complete the proof.

Remark 3. In the difinition, we can require that $U$ in condition 2 is homeomorphic to an open subset of $\R^n$. (Why? see, Boothby p.6)

Remark 4. Condition 1 is independent. Since we have the following example:

Example 1. Let $X=A_+\cup A_-\cup B$, $X\subset \R^2$, with
\begin{align*}
A_+&=\set{(x,y)|x\geq0,y=1},\\
A_-&=\set{(x,y)|x\geq0,y=-1},\\
B&=\set{(x,y)|x< 0, y=0}.
\end{align*}
Define the topology as follows, for the neighborhood of points other than $(0,\pm1)$, we use the induced topology from $\R^2$; and added $N^\pm_\eps=\set{(x,\pm1)|0\leq x< \eps}\cup\set{(x,0)|-\eps\leq x< 0}$ as neighborhood of $(0,\pm1)$.

It is a directly verification that the above definition do defined a topology for $X$ and it is locally Euclidean. Note that $X$ can’t be Hausdorff, since $(0,\pm1)$ can’t be separated by two non-intersection open set.

2. The Properties of Topological Manifolds Now, we’ll turn to the properties of topological manifolds.
Proposition 2. Suppose that $M$ is a Topological Manifolds, then
1. $M$ is locally connected, i.e., for every point $x\in M$, and any neighborhood $U$ of $x$, there is a open connected set $V$, such that $x\in V\subset U$;
2. Since $M$ is locally connected, then every connected component of $M$ is open;
3. If $M$ is connected, then it is connected by arc, i.e., for any two points $p,q\in M$, there exist an continuous map (called arc or path) $\phi:[0,1]\to M$, such that $\phi(0)=p, \phi(1)=q$;
4. $M$ is locally compact, i.e., for every point $x\in M$, and any neighborhood $U$ of $x$, there is a open set $V$ with $\overline V$ is compact, such that $x\in V\subset \overline V\subset U$;
5. Since every Hausdorff locally compact space is regular, so $M$ is regular. (Recall that a space is regular, if the point and closed set can be separated by their prospectively neighborhoods.)
The above property is not true for infinite dimensional space.
6. Manifold may not normal. (Recall that a space is normal if any two closed subset can be separated by their neighborhoods.)

Another important thing is about the second countability, we always need to assume that $M$ has a countable basis.
Proposition 3. For any topological manifold, the following properties are equivalent:
1. Each component of $M$ is $\sigma$-compact(also called compact at infinity), i.e., there exist a family of compact sets $\set{K_m}_{m=1}^\infty$, such that $K_{i-1}\subset K_i^\circ\subset K_{i+1}$ and $M=\cup_{m=1}^\infty K_m$;
2. Each component of $M$ is second-countable;
3. $M$ is metrizable;
4. $M$ is paracompact, i.e., each open covering has a locally finite refinement.
thus, any compact manifold is metrizable.