# General Topology/Connected spaces

Definition (connectedness):
Let $X$ be a topological space. $X$ is called connected if and only if whenever $U,V\subseteq X$ are two proper open subsets such that $U\cup V=X$, then $U\cap V\neq \emptyset$. A subset $S\subseteq X$ of a topological space is called connected if and only if it is connected with respect to the subspace topology.
Proposition (characterisation of connectedness):
Let $X$ be a topological space. The following are equivalent:
* $X$ is connected
* Whenever $\emptyset \subsetneq A,B\subsetneq X$ are closed so that $A\cup B=X$, then $A\cap B\neq \emptyset$
* $\emptyset$ and $X$ are the only subsets of $X$ which are simultaneously open and closed
Proof: If $X$ is connected, suppose that $A,B\subseteq X$ are closed so that $A\cup B=X$. Set $U:=X\setminus A$ and $V=X\setminus B$. Then $U\cap V=X\setminus (A\cup B)=\emptyset$, so that we find $y\in X\setminus (U\cup V)=A\cap B$. Suppose then that $S\subseteq X$ is clopen (ie. open and closed), and $S\notin \{\emptyset ,X\}$. Then $X=S\setminus (X\setminus S)$ is the disjoint union of two nontrivial closed subsets, contradiction. Finally, if $X=U\cup V$ and $V\cap U=\emptyset$, then $U$ and $V$ are both clopen. $\Box$
Example (two disjoint open balls in the real line are disconnected):
Consider the subspace $(0,1)\cup (2,3)$ of $\mathbb {R}$, equipped with the subspace topology. It is an example of a space which is not connected.
Proposition (continuous image of a connected space is connected):
Let $f:X\to Y$ be a continuous function, and suppose that $X$ is connected. Then $f(X)$ is connected with respect to its subspace topology (induced by $Y$).
Proof: Suppose that $U$ and $V$ are two open subsets of $f(X)$ such that $U\cup V=f(X)$ and $U\cap V=\emptyset$. By definition of the subspace topology, write $U=O\cap f(X)$ and $V=W\cap f(X)$, where $O,W$ are open in $Y$. Since $f$ is continuous, $f^{-1}(O)$ and $f^{-1}(W)$ are open in $X$. Further, $f^{-1}(O)\cap f^{-1}(W)=f^{-1}(O\cap W)=\emptyset$, since any element in $f^{-1}(O\cap W)$ would be mapped to $O\cap W\cap f(X)$. Finally, every element in $X$ is either mapped to $O$ or to $W$, so that $f^{-1}(O)\cup f^{-1}(W)=X$, and $X$ is not connected, a contradiction. $\Box$
Example (the closed unit interval is connected):
Set $X=[0,1]$ with the topology induced by the Euclidean topology on $\mathbb {R}$. Then $X$ is connected.
Proof: Let $U,V$ be two open subsets of $[0,1]$ so that $U\cap V=\emptyset$ and $U\cup V=X$. Suppose by renaming $U,V$ if necessary that $0\in U$. $V$ has an infimum, say $\eta \in \mathbb {R}$. Since $0\in U$, pick by openness of $U$ an $\epsilon >0$ such that $B_{\epsilon }(0)\subseteq U$. Then $B_{\epsilon }(0)\cap V=\emptyset$, so that $\eta >0$. Suppose that $\eta \in V$. Then $B_{\epsilon }(\eta )\subseteq V$ for some $\epsilon >0$, so that in particular $\eta -\epsilon /2\in V$, a contradiction to $\eta =\inf V$. Hence $\eta \in U$, but then pick $\epsilon >0$ so that $B_{\epsilon }(\eta )\subseteq U$ and obtain that $\inf V\geq \eta +\epsilon /2$, a contradiction. $\Box$
Definition (connected component):
Let $X$ be a topological space and let $x\in X$. The set of all $y\in X$ so that there exists $S\subseteq X$ which is connected and $x,y\in S$ is called the connected component of $x$.
Proposition (topological spaces decompose into connected components):
Let $X$ be a topological space. Then
    $X=\bigcup _{\alpha \in A}X_{\alpha }$,
where the union is disjoint and each $X_{\alpha }$ is the connected component of each of its points.
Proof: We prove that being contained within a common connected set is an equivalence relation, thereby proving that $X$ is partitioned into the equivalence classes with respect to that relation, thereby proving the claim. Indeed, it is certainly reflexive and symmetric. To prove it transitive, let $x,y\in S$ and $y,z\in T$, where $S$ and $T$ are connected. We claim that $S\cup T$ is connected; once this is proven, $x$ and $z$ will lie in a common connected set ($T\cup S$). Hence, let $U\cup V=S\cup T$, where $U,V$ are open with respect to the subspace topology on $S\cup T$, that is, $U=O\cap (S\cup T)$, $V=W\cap (S\cup T)$ for suitable $W,O$ that are open in $X$. Since $S$ is connected, $S\cap O=S$ or $S\cap W=S$ since $(S\cap O)\cup (S\cap W)=S$ and $(S\cap O)\cup (S\cap W)\subseteq U\cap V=\emptyset$. Suppose, by renaming $O$, $W$ if necessary, that $S\cap O=S$, that is, $S\subseteq O$. Note that by a similar argument, $T\cap O=T$ or $T\cap W=T$, but $T\cap W=T$ is impossible, since then $y\in W\cap O\cap (S\cup T)=U\cap V$. Hence, $T\cap O=T$ and $S\cup T\subseteq O$, so that $U=S\cup T$, and $S\cup T$ is connected. $\Box$
Definition (path):
Let $X$ be a topological space. A path is a continuous function $\gamma :[a,b]\to X$, where $a\leq b$.
Definition (concatenation of paths):
Let $\gamma :[a,b]\to X$ and $\rho :[c,d]\to X$ are two paths such that $\gamma (b)=\rho (c)$. Then the concatenation of $\gamma$ and $\rho$ is defined to be the path
    $\gamma *\rho :[0,1]\to X,\gamma *\rho (t):={\begin{cases}\gamma (2tb+(1-2t)a)&t\leq {\frac {1}{2}}\\\rho \left(2\left(t-{\frac {1}{2}}\right)d+2(1-t)c\right)\$t\geq {\frac {1}{2}}\end{cases}}$.
Proposition (concatenation of paths is continuous):
Let $\gamma :[a,b]\to X$ and $\rho :[c,d]\to X$ be two paths. Then $\gamma *\rho$ is continuous.
Proof: We have
    $\gamma *\rho |_{\left[0,{\frac {1}{2}}\right]}(t)=\gamma (2tb+(1-2t)a)$ and $\gamma *\rho |_{\left[{\frac {1,2}{,}}1\right]}(t)=\rho \left(2\left(t-{\frac {1}{2}}\right)d+2(1-t)c\right)$,
both of which are continuous. We conclude since a function continuous when restricted to two closed subsets which cover the space is continuous. $\Box$
Definition (path-connectedness):
Let $X$ be a topological space. $X$ is called path-connected if and only if for every two points $x,y\in X$, there exists a path $\gamma :[a,b]\to X$ such that $\gamma (a)=x$ and $\gamma (b)=y$. A subset $S\subseteq X$ is called path-connected iff, equipped with its subspace topology, it is a path-connected topological space.
That is, a space is path-connected if and only if between any two points, there is a path.
Proposition (path-connectedness implies connectedness):
Let $X$ be a path-connected topological space. Then $X$ is also connected.
Proof: Suppose that $X=U\cup V$, where $U,V$ are open and $U\cap V=\emptyset$. Suppose there exist $x\in U\setminus V$ and $y\in V\setminus U$, so that $U$ and $V$ are both proper nonempty subsets of $X$. Then consider by path-connectedness a path $\gamma :[a,b]\to X$ such that $\gamma (a)=x$ and $\gamma (b)=y$. $S:=\gamma ([a,b])$ is a connected subspace of $X$ since $[a,b]$ is a continuous image of the closed unit interval $[0,1]$ which is connected, and $\gamma ([a,b])$ is then connected as the continuous image of a connected set, since the continuous image of a connected space is connected. On the other hand, $(U\cap S)\cup (V\cap S)=X\cap S=S$ and $(U\cap S)\cap (V\cap S)\subseteq U\cap V=\emptyset$, where $(U\cap S)$ and $(V\cap S)$ are both open with respect to the subspace topology on $S$, so that $S$ is not connected, a contradiction. $\Box$
Proposition (connectedness by path is equivalence relation):
Let $X$ be any topological space. Then the relation
    $x\sim y:\Leftrightarrow \exists \gamma :[a,b]\to X{\text{ continuous }}:\gamma (a)=x,\gamma (b)=y$
is an equivalence relation.
Proof: For reflexivity, note that the constant function is always continuous. For symmetry, note that if we are given $\gamma :[a,b]\to X$ such that $\gamma (a)=x$ and $\gamma (b)=y$, we may consider the path
    ${\overline {\gamma }}:[a,b]\to X,{\overline {\gamma }}(t):=\gamma (a+b-t)$,
which is continuous as the composition of continuous functions and has the property that ${\overline {\gamma }}(0)=y$ and ${\overline {\gamma }}(1)=x$. Finally, whenever we have a path $\gamma :[a,b]\to X$ such that $\gamma (a)=x$ and $\gamma (b)=y$, and another path $\rho :[c,d]\to X$ such that $\rho (c)=y$ and $\rho (d)=z$, then $\gamma *\rho :[0,1]\to X$ is a path such that $\gamma *\rho (0)=x$ and $\gamma *rho(1)=z$, so that transitivity holds. $\Box$
Definition (path-connected component):
Let $X$ be a topological space, and let $x\in X$ be a point. The path-connected component of $x$ is the equivalence class of $x$, where $X$ is partitioned by the equivalence relation of path-connectedness.
Here we have a partial converse to the fact that path-connectedness implies connectedness:
Definition (local path-connectedness):
Let $X$ be a topological space. $X$ is called locally path-connected iff for every $x\in X$ and every open set $U$ of $x$ such that $x\in U$, there exists an open neighbourhood $V$ of $x$ such that $V\subseteq U$ which is path-connected.
Theorem (equivalence of connectedness and path-connectedness in locally path-connected spaces):
Let $X$ be a topological space which is locally path-connected. Then $X$ is connected if and only if it is path-connected.
Proof: First note that path-connected spaces are connected. Then suppose that $X$ is connected, fix $x_{0}\in X$, and define the set
    $S:=\{y\in X|\exists \gamma :[a,b]\to X:\gamma (a)=x_{0},\gamma (b)=y$.
Note that $S$ is open, since if $y\in S$, then by local path-connectedness we may pick a path-connected open neighbourhood $U$ of $y$, so that by applying concatenation, we see that all points in $U$ are in $S$. On the other hand, $X\setminus S$ is open, pretty much by the same argument: If $z\in X\setminus S$ and $U$ is a path-connected open neighbourhood of $x$, then $U\subseteq X\setminus S$, since if $U$ would contain a point $y$ of $S$, $x_{0}$ could be joined to $z$ by a path, concatenating a path from $x_{0}$ to $y$ to one from $y\to z$, in contradiction to $z\notin S$. Hence $S$ is open and closed, and since $x_{0}\in S$, $S\neq \emptyset$, so that $S=X$ by connectedness. $\Box$
This theorem has an important application: It proves that manifolds are connected if and only if they are path-connected. Also, later in this book we'll get to know further classes of spaces that are locally path-connected, such as simplicial and CW complexes.
By substituting "connected" for "path-connected" in the above definition, we get:
Definition (local connectedness):
Let $X$ be a topological space. $X$ is called locally connected if and only if for $x\in X$ and every neighbourhood $U$ of $x$, there exists a connected neighbourhood $V$ of $x$ such that $V\subseteq U$.

## Exercises

* Prove that whenever $X$ is a connected topological space and $Y$ is a topological space and $f:X\to Y$ is a continuous function, then $f(X)$ is connected with the subspace topology induced on it by $Y$.
* Prove that similarly if $X$ is a path-connected top. space, $Y$ top. space, $f:X\to Y$ continuous, then $f(X)$ is path-connected with the subspace topology induced on it by $Y$.
* Prove that a topological space $X$ is connected if and only if, when $\chi _{A}:X\to \{0,1\}$ for $A\subseteq X$ denotes the indicator function, the only indicator functions which are continuous are the ones where $A=\emptyset$ and $A=X$; here $\{0,1\}$ has the discrete topology.
* Let $X:=\{(0,0)\}\cup \{(x,y)|\exists n\in \mathbb {N} :y=1/n\}$ with the subspace topology induced by the Euclidean topology of $\mathbb {R} ^{2}$. Prove that $X$ is not locally connected by proving that $(0,0)$ does not have a connected neighbourhood.