# Topology/Path Connectedness

Topology
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## DefinitionEdit

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 ${\displaystyle f(1)=x_{1}}$

## ExampleEdit

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 ConnectednessEdit

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.

## ExercisesEdit

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 →