# Topology/Path Connectedness

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## Definition

A topological space ${\displaystyle X}$ is said to be path connected if for any two points ${\displaystyle x_{0},x_{1}\in X}$ there exists a continuous function ${\displaystyle f:[0,1]\to X}$ such that ${\displaystyle f(0)=x_{0}}$ and ${\displaystyle f(1)=x_{1}}$

## Example

1. All convex sets in a vector space are connected because one could just use the segment connecting them, which is ${\displaystyle f(t)=t{\vec {a}}+(1-t){\vec {b}}}$.
2. The unit square defined by the vertices ${\displaystyle [0,0],[1,0],[0,1],[1,1]}$ is path connected. Given two points ${\displaystyle (a_{0},b_{0}),(a_{1},b_{1})\in [0,1]\times [0,1]}$ the points are connected by the function ${\displaystyle f(t)=[(1-t)a_{0}+ta_{1},(1-t)b_{0}+tb_{1}]}$ for ${\displaystyle t\in [0,1]}$.
The preceding example works in any convex space (it is in fact almost the definition of a convex space).

Let ${\displaystyle X}$ be a topological space and let ${\displaystyle a,b,c\in X}$. Consider two continuous functions ${\displaystyle f_{1},f_{2}:[0,1]\to X}$ such that ${\displaystyle f_{1}(0)=a}$, ${\displaystyle f_{1}(1)=b=f_{2}(0)}$ and ${\displaystyle f_{2}(1)=c}$. Then the function defined by

${\displaystyle f(x)=\left\{{\begin{array}{ll}f_{1}(2x)&{\text{if }}x\in [0,{\frac {1}{2}}]\\f_{2}(2x-1)&{\text{if }}x\in [{\frac {1}{2}},1]\\\end{array}}\right.}$

Is a continuous path from ${\displaystyle a}$ to ${\displaystyle c}$. Thus, a path from ${\displaystyle a}$ to ${\displaystyle b}$ and a path from ${\displaystyle b}$ to ${\displaystyle c}$ can be adjoined together to form a path from ${\displaystyle a}$ to ${\displaystyle c}$.

## Relation to Connectedness

Each path connected space ${\displaystyle X}$ is also connected. This can be seen as follows:

Assume that ${\displaystyle X}$ is not connected. Then ${\displaystyle X}$ is the disjoint union of two open sets ${\displaystyle A}$ and ${\displaystyle B}$. Let ${\displaystyle a\in A}$ and ${\displaystyle b\in B}$. Then there is a path ${\displaystyle f}$ from ${\displaystyle a}$ to ${\displaystyle b}$, i.e., ${\displaystyle f:[0,1]\rightarrow X}$ is a continuous function with ${\displaystyle f(0)=a}$ and ${\displaystyle f(1)=b}$. But then ${\displaystyle f^{-1}(A)}$ and ${\displaystyle f^{-1}(B)}$ are disjoint open sets in ${\displaystyle [0,1]}$, covering the unit interval. This contradicts the fact that the unit interval is connected.

## Exercises

1. Prove that the set ${\displaystyle A=\{(x,f(x))|x\in \mathbb {R} \}\subset \mathbb {R} ^{2}}$, where ${\displaystyle f(x)=\left\{{\begin{array}{ll}0&{\text{if }}x\leq 0\\\sin({\frac {1}{x}})&{\text{if }}x>0\\\end{array}}\right.}$
is connected but not path connected.

 Topology ← Connectedness Path Connectedness Compactness →