# Is the product of path connected spaces also path connected in a topology other than the product topology?

In Munkres - Topology, in section 24, question 8a), we are asked "Is the product of path connected spaces necessarily path connected?" Here is my proof:

In the product topology, yes; since for any two points $$\mathbf{x}$$ and $$\mathbf{y}$$ in $$\prod_{\alpha\in J}X_{\alpha}$$ where $$J$$ is an index set and $$\{X_{\alpha}\}_{\alpha\in J}$$ is a family of path connected spaces, then for every $$x_{\alpha}$$ and $$y_{\alpha}$$ in $$X_{\alpha}$$ there is a path connecting them; call it $$f_{\alpha}'$$. Scale all these paths so they map from $$[0,1]$$ to $$X_{\alpha}$$, call the scaled paths $$f_{\alpha}$$. Then we can define the path $$\mathbf{x}$$ to $$\mathbf{y}$$ by $$f:[0,1]\longrightarrow\prod X_{\alpha}$$ with $$f(t)=(f_{\alpha}(t))_{\alpha\in J}$$. This function is guaranteed to be continuous (in the product topology) by theorem 19.6 (Munkres - Topology).

I think my proof is okay, but wouldn't mind it being looked over. However, I am unsure if perhaps I could have done a proof without relying on the topology being the product topology. I know that, for example, the uniform and box topologies on $$\mathbb{R}^{\omega}$$ are not even connected, so they can't be path connected. What about other topologies? What can we say in general about the product of connected (path connected or not) spaces under any particular topology?

• If we speak of the product of topological spaces, we mean the product topology. Of course, if you decide to endow the set $A\times B$ with the discrete topology, then the result will hardly ever be path-connected Commented May 27, 2018 at 20:09