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?

Thanks for reading.

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    $\begingroup$ 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 $\endgroup$ Commented May 27, 2018 at 20:09
  • $\begingroup$ @HagenvonEitzen I assumed as much, I was just curious if there were more general theorems applying to other topologies than the product topology. I know that anything coarser will also be path connected; is there anything that can be said about finer spaces beyond needing to actually look at that topology? $\endgroup$
    – Jay
    Commented May 27, 2018 at 21:17

1 Answer 1


Your proof for the product topology is fine: you use the important and basic characterisation of continuous maps that map into a product, and this characterisation of continuity actually uniquely determines the product topology, as I wrote about here, e.g. As path-connectedness is defined by the existence of certain continuous maps, the proof of path-connectedness is quite natural.

For a finer topology nothing sensible can really be said, and the first natural finer candidate, the box-topology, fails miserably: no non-trivial box-product is connected, let alone path-connected. Also other properties fail for finer topologies, compactness being a particularly important one. That's why the product topology is usually the only suitable candidate and its properties guarantee the path-connectedness of a product of path-connected spaces.


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