Exercise 2.4.7 (Every path-connected set is connected)

Proof. Let $E$ be a path-connected set. That is, for every $x,y\in E$ there is a continuous function $\gamma :[0,1]\to E$ such that $\gamma (0)=x$ and $\gamma (1)=y$.

Now suppose that $E$ is not connected. That is, there exists $V,W$ non-empty, disjoint, and open in $E$ such that $V\cup W=E$.

We will now show that $[0,1]$ is disconnected (which gives us a contradiction).

Since $\gamma$ is continuous and $V,W$ are open in $E$, we know that ${\gamma }^{-1}(V),{\gamma }^{-1}(W)$ are open in $[0,1]$.

Since $\gamma (0)=x$ and $\gamma (1)=y$ we have that both ${\gamma }^{-1}(V),{\gamma }^{-1}(W)$ are non-empty.

Suppose $z\in {\gamma }^{-1}(V)\cap {\gamma }^{-1}(W)$. That is, $\gamma (z)\in V$ and $\gamma (z)\in W$, but $V,W$ are disjoint, a contradiction. Hence ${\gamma }^{-1}(V)\cap {\gamma }^{-1}(W)=\varnothing$.

Let $z\in [0,1]$. Then $\gamma (z)\in E$, but $E=V\cup W$, and $V$, $W$ are disjoint. So either $\gamma (z)\in V\Rightarrow$ $z\in {\gamma }^{-1}(V)$ or $\gamma (z)\in W\Rightarrow z\in {\gamma }^{-1}(W)$. Thus $[0,1]={\gamma }^{-1}(V)\cup {\gamma }^{-1}(W)$.

Thus we have that $[0,1]$ is disconnected, a contradiction (Theorem 2.4.5.).

Hence our original assumption was wrong, and therefore every path-connected set is connected. □