# General Topology/Connected spaces

Definition (connectedness):

Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called connected if and only if whenever ${\displaystyle U,V\subseteq X}$ are two proper open subsets such that ${\displaystyle U\cup V=X}$, then ${\displaystyle U\cap V\neq \emptyset }$. A subset ${\displaystyle S\subseteq X}$ 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 ${\displaystyle X}$ be a topological space. The following are equivalent:

1. ${\displaystyle X}$ is connected
2. Whenever ${\displaystyle \emptyset \subsetneq A,B\subsetneq X}$ are closed so that ${\displaystyle A\cup B=X}$, then ${\displaystyle A\cap B\neq \emptyset }$
3. ${\displaystyle \emptyset }$ and ${\displaystyle X}$ are the only subsets of ${\displaystyle X}$ which are simultaneously open and closed

Proof: If ${\displaystyle X}$ is connected, suppose that ${\displaystyle A,B\subseteq X}$ are closed so that ${\displaystyle A\cup B=X}$. Set ${\displaystyle U:=X\setminus A}$ and ${\displaystyle V=X\setminus B}$. Then ${\displaystyle U\cap V=X\setminus (A\cup B)=\emptyset }$, so that we find ${\displaystyle y\in X\setminus (U\cup V)=A\cap B}$. Suppose then that ${\displaystyle S\subseteq X}$ is clopen (ie. open and closed), and ${\displaystyle S\notin \{\emptyset ,X\}}$. Then ${\displaystyle X=S\setminus (X\setminus S)}$ is the disjoint union of two nontrivial closed subsets, contradiction. Finally, if ${\displaystyle X=U\cup V}$ and ${\displaystyle V\cap U=\emptyset }$, then ${\displaystyle U}$ and ${\displaystyle V}$ are both clopen. ${\displaystyle \Box }$

Example (two disjoint open balls in the real line are disconnected):

Consider the subspace ${\displaystyle (0,1)\cup (2,3)}$ of ${\displaystyle \mathbb {R} }$, 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 ${\displaystyle f:X\to Y}$ be a continuous function, and suppose that ${\displaystyle X}$ is connected. Then ${\displaystyle f(X)}$ is connected with respect to its subspace topology (induced by ${\displaystyle Y}$).

Proof: Suppose that ${\displaystyle U}$ and ${\displaystyle V}$ are two open subsets of ${\displaystyle f(X)}$ such that ${\displaystyle U\cup V=f(X)}$ and ${\displaystyle U\cap V=\emptyset }$. By definition of the subspace topology, write ${\displaystyle U=O\cap f(X)}$ and ${\displaystyle V=W\cap f(X)}$, where ${\displaystyle O,W}$ are open in ${\displaystyle Y}$. Since ${\displaystyle f}$ is continuous, ${\displaystyle f^{-1}(O)}$ and ${\displaystyle f^{-1}(W)}$ are open in ${\displaystyle X}$. Further, ${\displaystyle f^{-1}(O)\cap f^{-1}(W)=f^{-1}(O\cap W)=\emptyset }$, since any element in ${\displaystyle f^{-1}(O\cap W)}$ would be mapped to ${\displaystyle O\cap W\cap f(X)}$. Finally, every element in ${\displaystyle X}$ is either mapped to ${\displaystyle O}$ or to ${\displaystyle W}$, so that ${\displaystyle f^{-1}(O)\cup f^{-1}(W)=X}$, and ${\displaystyle X}$ is not connected, a contradiction. ${\displaystyle \Box }$

Example (the closed unit interval is connected):

Set ${\displaystyle X=[0,1]}$ with the topology induced by the Euclidean topology on ${\displaystyle \mathbb {R} }$. Then ${\displaystyle X}$ is connected.

Proof: Let ${\displaystyle U,V}$ be two open subsets of ${\displaystyle [0,1]}$ so that ${\displaystyle U\cap V=\emptyset }$ and ${\displaystyle U\cup V=X}$. Suppose by renaming ${\displaystyle U,V}$ if necessary that ${\displaystyle 0\in U}$. ${\displaystyle V}$ has an infimum, say ${\displaystyle \eta \in \mathbb {R} }$. Since ${\displaystyle 0\in U}$, pick by openness of ${\displaystyle U}$ an ${\displaystyle \epsilon >0}$ such that ${\displaystyle B_{\epsilon }(0)\subseteq U}$. Then ${\displaystyle B_{\epsilon }(0)\cap V=\emptyset }$, so that ${\displaystyle \eta >0}$. Suppose that ${\displaystyle \eta \in V}$. Then ${\displaystyle B_{\epsilon }(\eta )\subseteq V}$ for some ${\displaystyle \epsilon >0}$, so that in particular ${\displaystyle \eta -\epsilon /2\in V}$, a contradiction to ${\displaystyle \eta =\inf V}$. Hence ${\displaystyle \eta \in U}$, but then pick ${\displaystyle \epsilon >0}$ so that ${\displaystyle B_{\epsilon }(\eta )\subseteq U}$ and obtain that ${\displaystyle \inf V\geq \eta +\epsilon /2}$, a contradiction. ${\displaystyle \Box }$

Definition (connected component):

Let ${\displaystyle X}$ be a topological space and let ${\displaystyle x\in X}$. The set of all ${\displaystyle y\in X}$ so that there exists ${\displaystyle S\subseteq X}$ which is connected and ${\displaystyle x,y\in S}$ is called the connected component of ${\displaystyle x}$.

Proposition (topological spaces decompose into connected components):

Let ${\displaystyle X}$ be a topological space. Then

${\displaystyle X=\bigcup _{\alpha \in A}X_{\alpha }}$,

where the union is disjoint and each ${\displaystyle X_{\alpha }}$ 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 ${\displaystyle X}$ 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 ${\displaystyle x,y\in S}$ and ${\displaystyle y,z\in T}$, where ${\displaystyle S}$ and ${\displaystyle T}$ are connected. We claim that ${\displaystyle S\cup T}$ is connected; once this is proven, ${\displaystyle x}$ and ${\displaystyle z}$ will lie in a common connected set (${\displaystyle T\cup S}$). Hence, let ${\displaystyle U\cup V=S\cup T}$, where ${\displaystyle U,V}$ are open with respect to the subspace topology on ${\displaystyle S\cup T}$, that is, ${\displaystyle U=O\cap (S\cup T)}$, ${\displaystyle V=W\cap (S\cup T)}$ for suitable ${\displaystyle W,O}$ that are open in ${\displaystyle X}$. Since ${\displaystyle S}$ is connected, ${\displaystyle S\cap O=S}$ or ${\displaystyle S\cap W=S}$ since ${\displaystyle (S\cap O)\cup (S\cap W)=S}$ and ${\displaystyle (S\cap O)\cup (S\cap W)\subseteq U\cap V=\emptyset }$. Suppose, by renaming ${\displaystyle O}$, ${\displaystyle W}$ if necessary, that ${\displaystyle S\cap O=S}$, that is, ${\displaystyle S\subseteq O}$. Note that by a similar argument, ${\displaystyle T\cap O=T}$ or ${\displaystyle T\cap W=T}$, but ${\displaystyle T\cap W=T}$ is impossible, since then ${\displaystyle y\in W\cap O\cap (S\cup T)=U\cap V}$. Hence, ${\displaystyle T\cap O=T}$ and ${\displaystyle S\cup T\subseteq O}$, so that ${\displaystyle U=S\cup T}$, and ${\displaystyle S\cup T}$ is connected. ${\displaystyle \Box }$

Definition (path):

Let ${\displaystyle X}$ be a topological space. A path is a continuous function ${\displaystyle \gamma :[a,b]\to X}$, where ${\displaystyle a\leq b}$.

Definition (concatenation of paths):

Let ${\displaystyle \gamma :[a,b]\to X}$ and ${\displaystyle \rho :[c,d]\to X}$ are two paths such that ${\displaystyle \gamma (b)=\rho (c)}$. Then the concatenation of ${\displaystyle \gamma }$ and ${\displaystyle \rho }$ is defined to be the path

${\displaystyle \gamma *\rho :[0,1]\to X,\gamma *\rho (t):={\begin{cases}\gamma (2tb+(1-2t)a)&t\leq {\frac {1}{2}}\\\rho \left(2\left(t-{\frac {1}{2}}\right)d+2(1-t)c\right)\t\geq {\frac {1}{2}}\end{cases}}}$.

Proposition (concatenation of paths is continuous):

Let ${\displaystyle \gamma :[a,b]\to X}$ and ${\displaystyle \rho :[c,d]\to X}$ be two paths. Then ${\displaystyle \gamma *\rho }$ is continuous.

Proof: We have

${\displaystyle \gamma *\rho |_{\left[0,{\frac {1}{2}}\right]}(t)=\gamma (2tb+(1-2t)a)}$ and ${\displaystyle \gamma *\rho |_{\left[{\frac {1,2}{,}}1\right]}(t)=\rho \left(2\left(t-{\frac {1}{2}}\right)d+2(1-t)c\right)}$,

both of which are continuous. We conclude since a function continuous when restricted to two closed subsets which cover the space is continuous. ${\displaystyle \Box }$

Definition (path-connectedness):

Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called path-connected if and only if for every two points ${\displaystyle x,y\in X}$, there exists a path ${\displaystyle \gamma :[a,b]\to X}$ such that ${\displaystyle \gamma (a)=x}$ and ${\displaystyle \gamma (b)=y}$. A subset ${\displaystyle S\subseteq X}$ 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 ${\displaystyle X}$ be a path-connected topological space. Then ${\displaystyle X}$ is also connected.

Proof: Suppose that ${\displaystyle X=U\cup V}$, where ${\displaystyle U,V}$ are open and ${\displaystyle U\cap V=\emptyset }$. Suppose there exist ${\displaystyle x\in U\setminus V}$ and ${\displaystyle y\in V\setminus U}$, so that ${\displaystyle U}$ and ${\displaystyle V}$ are both proper nonempty subsets of ${\displaystyle X}$. Then consider by path-connectedness a path ${\displaystyle \gamma :[a,b]\to X}$ such that ${\displaystyle \gamma (a)=x}$ and ${\displaystyle \gamma (b)=y}$. ${\displaystyle S:=\gamma ([a,b])}$ is a connected subspace of ${\displaystyle X}$ since ${\displaystyle [a,b]}$ is a continuous image of the closed unit interval ${\displaystyle [0,1]}$ which is connected, and ${\displaystyle \gamma ([a,b])}$ 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, ${\displaystyle (U\cap S)\cup (V\cap S)=X\cap S=S}$ and ${\displaystyle (U\cap S)\cap (V\cap S)\subseteq U\cap V=\emptyset }$, where ${\displaystyle (U\cap S)}$ and ${\displaystyle (V\cap S)}$ are both open with respect to the subspace topology on ${\displaystyle S}$, so that ${\displaystyle S}$ is not connected, a contradiction. ${\displaystyle \Box }$

Proposition (connectedness by path is equivalence relation):

Let ${\displaystyle X}$ be any topological space. Then the relation

${\displaystyle x\sim y:\Leftrightarrow \exists \gamma :[a,b]\to X{\text{ continuous }}:\gamma (a)=x,\gamma (b)=y}$

is an equivalence relation.

Proof: For reflexivity, note that the constant function is always continuous. For symmetry, note that if we are given ${\displaystyle \gamma :[a,b]\to X}$ such that ${\displaystyle \gamma (a)=x}$ and ${\displaystyle \gamma (b)=y}$, we may consider the path

${\displaystyle {\overline {\gamma }}:[a,b]\to X,{\overline {\gamma }}(t):=\gamma (a+b-t)}$,

which is continuous as the composition of continuous functions and has the property that ${\displaystyle {\overline {\gamma }}(0)=y}$ and ${\displaystyle {\overline {\gamma }}(1)=x}$. Finally, whenever we have a path ${\displaystyle \gamma :[a,b]\to X}$ such that ${\displaystyle \gamma (a)=x}$ and ${\displaystyle \gamma (b)=y}$, and another path ${\displaystyle \rho :[c,d]\to X}$ such that ${\displaystyle \rho (c)=y}$ and ${\displaystyle \rho (d)=z}$, then ${\displaystyle \gamma *\rho :[0,1]\to X}$ is a path such that ${\displaystyle \gamma *\rho (0)=x}$ and ${\displaystyle \gamma *rho(1)=z}$, so that transitivity holds. ${\displaystyle \Box }$

Definition (path-connected component):

Let ${\displaystyle X}$ be a topological space, and let ${\displaystyle x\in X}$ be a point. The path-connected component of ${\displaystyle x}$ is the equivalence class of ${\displaystyle x}$, where ${\displaystyle X}$ 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 ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called locally path-connected iff for every ${\displaystyle x\in X}$ and every open set ${\displaystyle U}$ of ${\displaystyle x}$ such that ${\displaystyle x\in U}$, there exists an open neighbourhood ${\displaystyle V}$ of ${\displaystyle x}$ such that ${\displaystyle V\subseteq U}$ which is path-connected.

Theorem (equivalence of connectedness and path-connectedness in locally path-connected spaces):

Let ${\displaystyle X}$ be a topological space which is locally path-connected. Then ${\displaystyle X}$ is connected if and only if it is path-connected.

Proof: First note that path-connected spaces are connected. Then suppose that ${\displaystyle X}$ is connected, fix ${\displaystyle x_{0}\in X}$, and define the set

${\displaystyle S:=\{y\in X|\exists \gamma :[a,b]\to X:\gamma (a)=x_{0},\gamma (b)=y}$.

Note that ${\displaystyle S}$ is open, since if ${\displaystyle y\in S}$, then by local path-connectedness we may pick a path-connected open neighbourhood ${\displaystyle U}$ of ${\displaystyle y}$, so that by applying concatenation, we see that all points in ${\displaystyle U}$ are in ${\displaystyle S}$. On the other hand, ${\displaystyle X\setminus S}$ is open, pretty much by the same argument: If ${\displaystyle z\in X\setminus S}$ and ${\displaystyle U}$ is a path-connected open neighbourhood of ${\displaystyle x}$, then ${\displaystyle U\subseteq X\setminus S}$, since if ${\displaystyle U}$ would contain a point ${\displaystyle y}$ of ${\displaystyle S}$, ${\displaystyle x_{0}}$ could be joined to ${\displaystyle z}$ by a path, concatenating a path from ${\displaystyle x_{0}}$ to ${\displaystyle y}$ to one from ${\displaystyle y\to z}$, in contradiction to ${\displaystyle z\notin S}$. Hence ${\displaystyle S}$ is open and closed, and since ${\displaystyle x_{0}\in S}$, ${\displaystyle S\neq \emptyset }$, so that ${\displaystyle S=X}$ by connectedness. ${\displaystyle \Box }$

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 ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called locally connected if and only if for ${\displaystyle x\in X}$ and every neighbourhood ${\displaystyle U}$ of ${\displaystyle x}$, there exists a connected neighbourhood ${\displaystyle V}$ of ${\displaystyle x}$ such that ${\displaystyle V\subseteq U}$.

## Exercises

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