A space \(X\) is locally connected if for each \(x\in X\) and each open set \(U\) containing \(x\), there is a connected open set \(V\) in \(X\) satisfying \(x\in V\subseteq U\). Let \(X\) and \(Y\) be locally connected spaces. Determine whether or not the product space \(X\times Y\) must also be locally connected.
Solution
Yes: \(X\times Y\) must be locally connected.
1. If two connected subsets \(C,D\) of a space have nonempty intersection, then \(C\cup D\) is connected.
Suppose \(C\cup D\) had a separation \[ C\cup D=U\cup V \] into disjoint nonempty relatively open sets. Connectedness of \(C\) forces \(C\) to lie entirely in one of \(U,V\), and connectedness of \(D\) forces the same for \(D\). Because \(C\cap D\ne\varnothing\), the two sets cannot lie in different members of the separation. Thus both lie in the same one, making the other empty, a contradiction.
2. The product of two nonempty connected spaces is connected.
Let \(C,D\) be nonempty and connected, and fix \[ c_0\in C, \qquad d_0\in D. \] The subset \[ C\times\{d_0\} \] is homeomorphic to \(C\), hence connected. For each \(c\in C\), the subset \[ \{c\}\times D \] is homeomorphic to \(D\), hence connected, and it meets \(C\times\{d_0\}\) at \((c,d_0)\). Therefore 1 implies that \[ E_c=(C\times\{d_0\})\cup(\{c\}\times D) \] is connected.
Every \(E_c\) contains the common connected subset \(C\times\{d_0\}\). If their union were separated, each connected \(E_c\) would have to lie in one member of the separation, and the common subset would force every \(E_c\) into the same member. Hence the union is connected. But \[ \bigcup_{c\in C}E_c=C\times D. \] Thus \(C\times D\) is connected.
3. Let \((x_0,y_0)\in X\times Y\) and let \(W\) be an open neighborhood of \((x_0,y_0)\).There are open sets \(U\subseteq X\) and \(V\subseteq Y\) such that\[ (x_0,y_0)\in U\times V\subseteq W. \]
The products of open sets form a basis for the product topology. Since \(W\) is open and contains \((x_0,y_0)\), it contains a basic product neighborhood of that point.
4. There are connected open sets \(U_0\subseteq X\) and \(V_0\subseteq Y\) such that\[ x_0\in U_0\subseteq U, \qquad y_0\in V_0\subseteq V. \]
Apply local connectedness of \(X\) to the open neighborhood \(U\) of \(x_0\), and local connectedness of \(Y\) to the open neighborhood \(V\) of \(y_0\).
5. The set \(U_0\times V_0\) is a connected open neighborhood of \((x_0,y_0)\) contained in \(W\).
4. The same step and 3 give\[ U_0\times V_0\subseteq U\times V\subseteq W. \]Both \(U_0\) and \(V_0\) are nonempty and connected, so 2 shows that their product is connected.
6. Therefore \(X\times Y\) is locally connected.
The point \((x_0,y_0)\) and its open neighborhood \(W\) in 3 were arbitrary. By 5, every such neighborhood contains a connected open neighborhood of the point. This is exactly the definition of local connectedness.