Topology/Path Connectedness

Definition

A topological space $X$ is said to be path connected if for any two points $x_0,x_1\in X$ there exists a continuous function $f:[0,1]\to X$ such that $f(0)=x_0$ and $f(1)=x_1$

Example

1. All convex sets in a vector space are connected because one could just use the segment connecting them, which is f(t)=ta+(1-t)b.
2. The unit square defined by the vertices $[0,0],[1,0],[0,1],[1,1]$ is path connected. Given two points $(a_0,b_0),(a_1,b_1)\in [0,1]\times [0,1]$ the points are connected by the function $f(t)=[(1-t)a_0+t{a}_1,(1-t)b_0+t{b}_1]$ for $t\in [0,1]$.
   The preceding example works in any convex space (it is in fact almost the definition of a convex space).

Adjoining Paths

Consider two continuous functions f₁ and f₂ defined within [0,1] so that f₁(0)=a, f₁(1)=b=f₂(0) and f₂(1)=c. Then the function defined by

$f(x)=\{\begin{array}{cc}{f}_1(2x)& \text{if }x\in [0,\frac{1}{2}]\\ f_2(2x-1)& \text{if }x\in [\frac{1}{2},1]\end{array}$

Is a continuous path from a to c. Thus, a path from a to b and a path from b to c can be adjoined together to form a path from a to c.

Relation to Connectedness

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

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

Exercises

1. Prove that the function $f(x)=\{\begin{array}{cc}0& \text{if }x\le 0\\ sin(\frac{1}{x})& \text{if }x>0\end{array}$
   is connected but not path connected.