In this post, I would like to explore some basic properties of Topological manifolds.<\/p>\n
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 \\emph{locally Euclidean} will make on the original topological space.
\n\\section{The Definition of Topological Manifolds}
\nLet’s first give the definition of \\emph{Topological Manifolds}.
\n\\begin{defn}
\nSuppose that $M$ is a topological space. If
\n\\begin{enumerate}
\n\t\\item $M$ 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 $U\\cap V=\\emptyset$.
\n\t\\item For each $x\\in M$, there is a neighbourhood $U$ of $x$ and an integer $n\\geq0$, such that $U$ is homeomorphic to $\\R^n$.
\n\\end{enumerate}
\nThen, we call that $M$ is a \\emph{Topological Manifolds}.
\n\\end{defn}
\n
\n\\begin{rem}
\ncondition 2 also called “\\emph{Locally Euclidean}”.
\n\\end{rem}
\n\\begin{rem}
\nIn condition 2, the neighborhood can be replaced by \\emph{open neiborhood of $M$}.
\n\\end{rem}
\n\\begin{proof}
\nSince 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 $x$.<\/p>\n
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: \\emph{If $\\phi$ is a homeomorphic between the topological space $M$ and $N$, and $U$ is a subset of $M$, then under the induced topology<\/em>, $\\phi$ 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. 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. In this post, I would like to explore so …<\/p>\n
\n\\end{proof}
\n\\begin{rem}
\nIn the difinition, we can require that $U$ in condition 2 is homeomorphic to an open subset of $\\R^n$. (Why? see, Boothby p.6)
\n\\end{rem}
\n\\begin{rem}
\nCondition 1 is independent. Since we have the following example:
\n\\end{rem}
\n\\begin{examp}
\nLet $X=A_+\\cup A_-\\cup B$, $X\\subset \\R^2$, with
\n\\begin{align*}
\nA_+&=\\set{(x,y)|x\\geq0,y=1},\\\\
\nA_-&=\\set{(x,y)|x\\geq0,y=-1},\\\\
\nB&=\\set{(x,y)|x< 0, y=0}.
\n\\end{align*}
\nDefine 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 $(0,\\pm1)$.<\/p>\n
\n\\end{examp}
\n\\section{The Properties of Topological Manifolds}
\nNow, we’ll turn to the properties of topological manifolds.
\n\\begin{prop}
\nSuppose that $M$ is a Topological Manifolds, then
\n\\begin{enumerate}
\n\\item $M$ 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 $x\\in V\\subset U$;
\n\\item Since $M$ is locally connected, then every connected component of $M$ is open;
\n\\item 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 $\\phi(0)=p, \\phi(1)=q$;
\n\\item $M$ 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 $x\\in V\\subset \\overline V\\subset U$;
\n\\item 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.)
\n\\emph{The above property is not true for infinite dimensional space}.
\n\\item Manifold may not normal. (Recall that a space is normal if any two closed subset can be separated by their neighborhoods.)
\n\\end{enumerate}
\n\\end{prop}
\nAnother important thing is about the second countability, we always need to assume that $M$ has a countable basis.
\n\\begin{prop}
\nFor any topological manifold, the following properties are equivalent:
\n\\begin{enumerate}
\n\\item 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 $M=\\cup_{m=1}^\\infty K_m$;
\n\\item Each component of $M$ is second-countable;
\n\\item $M$ is metrizable;
\n\\item $M$ is paracompact<\/a>, i.e., each open covering has a locally finite refinement.
\n\\end{enumerate}
\nthus, any compact manifold is metrizable.
\n\\end{prop}<\/p>\n","protected":false},"excerpt":{"rendered":"