Path Connectivity of Connected Topological Spaces

Recall from the Path Connected Topological Spaces page that a topological space $X$ is said to be path connected if for each pair of points $x, y \in X$ there exists a continuous function $\alpha : [0, 1] \to X$, call a path, such that $\alpha(0) = x$ and $\alpha(1) = y$.

Interestingly enough, path connectivity is actually a stronger concept that regular connectivity. As we will show in the following theorem, every path connected topological space is also a connected topological space.

Theorem 1: If $X$ is a path connected topological space then $X$ is also a connected topological space.

• Proof: Let $X$ be a path connected topological space and assume that instead $X$ is disconnected. Since $X$ is disconnected, there exists $A, B \subset X$, where $A, B \neq \emptyset$, $A \cap B = \emptyset$ and:

(1)

\begin{align} \quad X = A \cup B \end{align}

• Let $a \in A$ and $b \in B$. Since $X$ is path connected there exists a continuous function $\alpha : [0, 1] \to X$ such that $\alpha(0) = a$ and $\alpha(1) = b$. Since $\alpha$ is continuous, $\alpha^{-1}(A)$ and $\alpha^{-1}(B)$ are both open in $I$. We claim that $\{ \alpha^{-1}(A), \alpha^{-1}(B) \}$ is a separation of $I$.

• We have already established that $\alpha^{-1}(A)$ and $\alpha^{-1}(B)$ are open in $[0, 1]$. Furthermore, since $0 \in \alpha^{-1}(A)$ and $1 \in \alpha^{-1}(B)$ we see that $\alpha^{-1}(A), \alpha^{-1}(B) \neq \emptyset$. Since $X = A \cup B$ we see that:

(2)

\begin{align} \quad [0, 1] = \alpha^{-1}(X) = \alpha^{-1}(A \cup B) = \alpha^{-1}(A) \cup \alpha^{-1}(B) \end{align}

• We only need to show that $\alpha^{-1}(A) \cap \alpha^{-1}(B) = \emptyset$. Suppose not. Then there exists an $x \in \alpha^{-1}(A) \cap \alpha^{-1}(B)$, so $\alpha(x) \in A \cap B$. But $A \cap B = \emptyset$, which is a contradiction. Therefore $\alpha^{-1}(A) \cap \alpha^{-1}(B) = \emptyset$, so $\{ \alpha^{-1}(A), \alpha^{-1}(B) \}$ is a separation of $[0, 1]$. But $[0, 1]$ is a connected topological space (with the subspace topology), so we have arrived at a contradiction.

• Therefore the assumption that $X$ is disconnected was false. So if $X$ is path connected, $X$ is also connected. $\blacksquare$