Topology/Path Connectedness

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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


  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)a_0+ta_1,(1-t)b_0+tb_1] for t\in[0,1].

Adjoining Paths[edit]

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 f_1(0)=a, f_1(1)=b=f_2(0) and f_2(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[edit]

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.


  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.

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