\chapter{Connected spaces}


\section{Connected space}

Homeomorphisms can be used to show 2 topological spaces are the same, however it is interesting to note what properties distinguish topological spaces.

Consider the topological subspaces of the Euclidean topological space $A=[0,1]$ and $B=[0,1] \cup [2,3]$. They have a clear distinction; the former is a uniform interval while the latter has a gap in between.

We see that $B$ is comprised of 2 closed intervals 1 unit of distance away from another, but topologies cannot work in terms of distance. However the topology can detect that there are 2 disjoint open sets that split $B$ up.

Formally, the idea of connectedness captures the property of a topological space where it cannot be broken down into disjoint open sets.

We can consider the connectedness of entire spaces, and then extend this notion to general sets by considering the topological subspaces they induce (analogous to the above example).




\begin{definition}
A \emph{connected space} $(X,\mathcal{T})$ is a topological space such that the only union of a pair of disjoint open sets is the space with the emptyset. There are no disjoint open sets $U,V$ (other than $\empty,X$) where $U \cup V = X$.
\end{definition}



The following theorem provides an alternative but equivalent definition of connected spaces that is often easier to work with.

\begin{theorem}[Equivalent definitions of a connected space]
	Let $(X,\mathcal{T})$ be a topological space. $(X,\mathcal{T})$ is connected space iff any of the following hold.
\begin{itemize}
	\item There are no disjoint open sets $U,V$ (other than $\empty,X$) where $U \cup V = X$.
	\item There are no disjoint closed sets $U,V$ (other than $\empty,X$) where $U \cup V = X$.
	\item $\emptyset,X$ are the only clopen sets in $(X,\mathcal{T})$
\end{itemize}
\end{theorem}


Using the least upper bound property, the following many be proven.
\begin{proposition}
Euclidean topological space is connected.
\end{proposition}


\subsection{Connected sets}


\begin{definition}[Connected set]
Let $(X,\mathcal{T})$ be a topological space, then $S$ is a \emph{connected set of $X$} iff its subspace topology induced by $X$ is a connected space.
\end{definition}

Our knowlegde of topological subspaces comes in clutch here.

\begin{proposition}
Let $(X,\mathcal{T})$ be a topological space and $X=U \cup V$ a disjoint union of open sets. All connected sets are a subset of either $U$ or $V$.
\end{proposition}


Given a connected set $S$, extending $S$ by including limit points preserves its connectedness since the neighborhoods of limit points are never disjoint from $S$ ()

\begin{proposition}
Let $(X,\mathcal{T})$ be a topological space and $S$ a connected set of $X$, then any set $T$ satisfying $S \subseteq T \subseteq \mathrm{cl}(S)$ is a connected set.
\end{proposition}


\begin{proposition}
Let $(X,\mathcal{T})$ be a topological space, a set $S$ is connected iff $S$ has a nontrivial subset with empty boundary
	\[ S \text{ is connected } \iff \forall T \subseteq S [ \partial_S (T) \neq \emptyset ] \]
\end{proposition}

\begin{proposition}
\[ S,T \text{ are connected } \]
\[ S\cup T \text{ is connected } \iff \exists p \in S \cup T [ \{ p \} \cup S \text{ is connected } \land \{ p \} \cup T \text{ is connected } ]  \]
\end{proposition}

\begin{proposition}
	\[ f : X \to Y \text{ is continuous } \land C \text{ is connected } \implies f(C) \text{ is connected } \]
\end{proposition}




\begin{proposition}
Let $(X,\mathcal{T})$ be a topological space and $(S_i)_{i \in I}$ a family of connected sets such that $\bigcap_{i \in I} S_i \neq \emptyset$, then $\bigcup_{i\in I} S_i$ is a connected set
\end{proposition}

\begin{proposition}
finite box topology of connected spaces is connected
\end{proposition}


In real analysis, there are many theorems that only apply to connected sets, namely the intermediate value theorem and the fact that $f'(x)=0$ for all $x$ on a connected set (i.e an interval) implies $f$ is constant on said set.



\begin{proposition}[A fixed point theorem]
Let $f : [0,1] \to [0,1]$ be a continuous function, then $f$ has a fixed point.
\end{proposition}

\section{Path connected space}

An even more natural idea of connectedness is if any pair of points in the topological space can be, well, connected!

\begin{definition}[Curve]
A \emph{curve} is a continuous function $f : I \to X$, where $I$ is a non-degenerate interval of $\mathbb{R}$.
\end{definition}

\begin{definition}[Path]
A \emph{path} is a curve $f : [0,1] \to X$ with its domain as $[0,1]$
\end{definition}

We could have defined a path with any compact interval domain, but all compact intervals are homeomorphic, so $[0,1]$ was chosen for simplicity; a change of variables can make any curve with compact onterval domain a path.
\begin{definition}
A \emph{path connected space} $(X,\mathcal{T})$ is a topological space such that for any pair of distinct points $x,y \in X$, there exists a continuous function $f : [0,1] \to X$ where $f(0)=x$ and $f(1)=y$. Essentially, any two points in a space has a continuous path from one another.
\end{definition}

\begin{proposition}
Path connected spaces are connected spaces.
\end{proposition}

It's notable that connected spaces aren't always path connected; there is a famous counterexample called the \emph{Topologist's sine curve} that demonstrates this by being a discontinuous function that can still pass as connected.
Consider the following topological space as this graph with the induced Euclidean topology $\mathbb{R}^2$.
\[T=\{ \begin{bmatrix}x \\ \sin (\frac{1}{x})\end{bmatrix} : x \in (0,1] \} \cup \{\mathbf{0}\}\]

Note that no path can go through $\mathbf{0}$, since this would not be a continuous function and hence not a path (or even a curve). $\sin (\frac{1}{x})$ just oscillates like crazy near the origin; $\sin(\frac{1}{x})$ doesn't get closer to $0$ as $x$ approaches 0, therefore we looe continuity here and hence it is not a path connected space. That said, $T$ is still connected!



There is also the notion of a simply connected space; this is useful in the study of complex analysis, however to give a propert topological treatment of this property we require the methods of algebraic topology.


\section{Locally connected spaces}