If \(X\) is a path connected space, then \(X\) is connected.
Solution
1. The interval \([0,1]\) is connected.
Suppose instead that \[ [0,1]=U\cup V \] with \(U\) and \(V\) disjoint nonempty sets open in the subspace topology. Choose \(u\in U\) and \(v\in V\) with \(u<v\) after exchanging the names if necessary. Let \[ s=\sup\bigl(U\cap[u,v]\bigr). \] Then \(u\le s\le v\).
If \(s\in U\), openness of \(U\) gives an interval around \(s\) contained in \(U\) relative to \([0,1]\); since \(s<v\), this produces a point of \(U\cap[u,v]\) larger than \(s\), contradicting the definition of \(s\). Thus \(s\notin U\), so \(s\in V\). Openness of \(V\) gives an interval around \(s\) contained in \(V\) relative to \([0,1]\). By the definition of supremum, there is a point of \(U\cap[u,v]\) arbitrarily close to \(s\) from below, hence inside that interval, contradicting \(U\cap V=\varnothing\). Therefore no such separation exists.
2. Suppose for contradiction that the path-connected space \(X\) is disconnected.Then there are disjoint nonempty open sets \(U,V\subseteq X\) with\[ X=U\cup V. \]
This is the definition of disconnectedness.
3. Choose \(x\in U\) and \(y\in V\), and let\[ \gamma:[0,1]\to X \]be a path from \(x\) to \(y\).
The sets \(U\) and \(V\) are nonempty, and path connectedness supplies a path between any two points of \(X\).
4. The sets \(\gamma^{-1}(U)\) and \(\gamma^{-1}(V)\) form a separation of \([0,1]\).
They are open because \(\gamma\) is continuous. They are disjoint because \(U\) and \(V\) are disjoint, and together they cover \([0,1]\) because \(U\cup V=X\). They are both nonempty since \[ 0\in\gamma^{-1}(U), \qquad 1\in\gamma^{-1}(V). \]
5. This contradicts 1, so \(X\) is connected.
The separation in 4 cannot exist because \([0,1]\) is connected.