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\begin{document}
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\section*{IV.2 Basic topological properties}
\subsection* {4. Connectedness }
\begin{defn}
A topological space $X$ is disconnected if there is nonempty open
sets $A$ and $B$ in $X$
s.t. $X=A\cup B$ and $A\cap B=\phi$.\\
In this case $\{A,B\}$ is called a disconnection or a separation of $X$.\\
A topological space $X$ is connected if it is not disconnected.
\end{defn}
\begin{ex}
$\mathbb{Q}$ is disconnected by $(-\infty,\surd\overline{2})$ and
$(\surd\overline{2},\infty)$.
\end{ex}
\begin{prop}
$\re$ is connected.
\end{prop}
\begin{proof}
Suppose not. Then $\re=A\cup B, A\cap B=\phi$ where $A$ and
$B$ are open.\\
Choose $a\in A,\, b\in B$ and we may assume that $a**0)$ (topologist's
sine curve) and $\{0\}\times[-1,1]$ is the closure of the graph and
hence it is connected.
\end{ex}
\subsection* {6. Path-connectedness }
\begin{defn}
Let $X$ be a topological space. A continuous map
$\gamma:I=[0,1]\rightarrow X$ is called a path
joining $\gam(0)$ and $\gam(1)$.\\
A space $X$ is path-connected if each pair of points can be joined
by a path.
\end{defn}
\begin{prop}
Path-connected $\Rightarrow$ connected. (not $\Leftarrow$)
\end{prop}
\begin{proof}
Let $f:X\rightarrow \{0,1\}$, where $X$ is path-connected, be a
continuous function. If there exists $x$ and $y$ s.t. $f(x)=0$ and
$f(y)=1$, then there is
$\gam:I\rightarrow X$ s.t. $\gam(0)=x, \gam(1)=y$\\
$\Rightarrow f\circ\gamma :I\rightarrow\{0,1\}$ continuous and onto.\\
$\Rightarrow$ $I$ is not connected. (A contradiction!)\\\\
A counterexample of ($\Leftarrow$) : the closure of the topologist's
sine curve
\end{proof}
\begin{rem}
(1) Let $A_\alp$ be path connected for all $\alp\in I$. Then
$\cap_{\alp\in I} A_{\alp} \neq\phi
\Rightarrow \cup_{\alp\in I}A_{\alp}$ is path connected.\\
(2) The closure of a path connected space is not necessarily path
connected.(A counterexample is the topologist's sine curve.)
\end{rem}
\begin{defn}
A maximal (path-) connected subspace of a topological space is
called a (path-) component of the space.
\end{defn}
[Figure describing component and path-component on real line,
topologist' sine curve, rational number, and Cantor set]
\begin{defn}
A topological space is said to be totally disconnected if every
component is a point.
\end{defn}
\begin{prop}
$X$ is a topological space.\\
(1) Each point in $X$ is contained in exactly one (path-)
component of $X$.\\
(2) $X$ is a disjoint union of (path-) components.\\
(3) Each component is closed. (not necessarily for path-component)
\end{prop}
\begin{proof}
(1) The union of all (path-)connected sets containing $x\in X$ is
a (path-)component.\\
(2) If they intersect, their union will be connected, so it is a
contradiction to the maximality of (path-)component. Therefore
components are disjoint.\\
(3) If $C$ is a component. Then $C$ is connected and so is
$\overline{C}$. By maximality of component, $C\supset\overline{C}$.
Therefore $C=\overline{C}$ is closed.\\
(3) does not hold for path component. (A counterexample is the
topologist's sine curve).
\end{proof}
\begin{prop}
(1) $\forall x\in X$ has a (path-) connected
neighborhood\\
$\Leftrightarrow\forall$ each (path-)component
is open(and hence closed).\\
(2) $X$ is path-connected\\
$\Leftrightarrow$ $X$ is connected and $\forall x\in X$ has a
path-connected neighborhood.
\end{prop}
\begin{proof}
(1)($\Rightarrow$) Let $C$ be a component. Then by maximality,
$U_x\subset
C$ ($U_x$ is a connected neighborhood of $x$).\\
($\Leftarrow$) Trivial. \\
By the same argument this is true for path component, too.\\
(2)($\Rightarrow$) Clear.\\
($\Leftarrow$) By (1), each path-component is open. So they are
disjoint and open. Therefore each path-component is closed since its
complement is open being a disjoint union of open sets. Hence it is
both open and closed. Since $X$ is connected, a path-component
becomes $X$ itself.
\end{proof}
\begin{cor}
An open set in $\re^n$ is connected $\Leftrightarrow$ it is path-connected.
\end{cor}
\begin{defn}
A space $X$ is locally (path-) connected if for all $x$ in $X$, each
neighborhood of $x$ contains a (path) connected neighborhood.(i.e.,
each point has a basis consisting of connected open sets).
\end{defn}
\begin{rem}
Concepts of connectedness and local connectedness are independent,
i.e., one does not necessarily imply the other.
\end{rem}
Indeed an example of locally connected but not connected space is
$(0,1)\cup
(2,3)$.\\
An example of connected but not locally connected is the closure of
topologist sine curve.
\begin{prop}
(1) $X$ is locally (path-)connected \\
$\Leftrightarrow$ the (path-)components of each open set are open.\\
(In particular, each (path-) component is open for a locally
(path-)connected space).\\
(2) $X$ is locally path-connected\\
$\Rightarrow$ the components and the path-components of $X$ are the
same.
\end{prop}
Proof is a Homework.
\end{document}
**