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) = x₀ and f(1) = x₁

Example

1. All convex sets in a vector space are connected because one could just use the segment connecting them, which is $f(t)=t\vec{a}+(1-t)\vec{b}$.
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)a₀ + ta₁,(1 − t)b₀ + tb₁] 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

Let X be a topological space and let $a,b,c\in X$. Consider two continuous functions $f_1,f_2:[0,1]\to X$ such that f₁(0) = a, f₁(1) = b = f₂(0) and f₂(1) = c. Then the function defined by

$f(x) = \left\{ \begin{array}{ll} f_1(2x) & \text{if } x \in [0,\frac{1}{2}]\\ f_2(2x-1) & \text{if } x \in [\frac{1}{2},1]\\ \end{array} \right.$

Is a continuous path from 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., $f:[0,1]\rightarrow X$ 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 set $A=\{(x,f(x))|x\in\mathbb{R}\}\subset\mathbb{R}^2$, where $f(x) = \left\{ \begin{array}{ll} 0 & \text{if } x \leq 0\\ \sin(\frac{1}{x}) & \text{if } x > 0\\ \end{array} \right.$
   is connected but not path connected.