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Let $\{X_\alpha\}_{\alpha \in J}$ be a collection of connected topological spaces, where the index set $J$ is uncountable. How can we determine whether the cartesian product of these spaces is connected or not in the product topology or in the box topology?

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Maybe you can use that $X$ is connected if and only if every continuous function $X \to \{0,1\}$ is constant. –  Damien L Feb 10 '13 at 9:29
    
While I think I have come across this assertion somewhere, I would appreciate if you could please give a proof thereof too. And also how to answer my query using this assertion. –  Saaqib Mahmuud Feb 10 '13 at 9:34
    
If $f:X\to\{0,1\}$ is continuous and surjective, then $f^{-1}\big[\{0\}\big]$ and $f^{-1}\big[\{1\}\big]$ are disjoint, non-empty clopen sets whose union is $X$, and $X$ is not therefore connected. Conversely, if $X$ is not connected, then $X=U\cup V$ for some non-empty, disjoint, clopen sets $U,V$, and the function that sends every $x\in U$ to $0$ and every $x\in V$ to $1$ is continuous. This proves @Damien’s assertion. It’s not clear that it’s especially useful here, however. –  Brian M. Scott Feb 10 '13 at 9:39
    
I'm really sorry, but I just don't know how to accept an answer. My apologies! –  Saaqib Mahmuud Feb 10 '13 at 10:12

1 Answer 1

The box product of infinitely many non-trivial Tikhonov spaces is never connected; this is Theorem 1.3(iii) of Scott W. Williams, Box Products, in the Handbook of Set-Theoretic Topology, K. Kunen & J.E. Vaughan, eds., North-Holland, 1984. An arbitrary Tikhonov product of connected spaces, however, is always connected; you’ll find a proof here.

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Are you sure you're stating Williams' result correctly? If X is [0,1] under the cofinite topology, any two nonempty open sets intersect, so if we take the infinite box product, any two nonempty open boxes intersect; thus the product is connected (since it can't be partitioned into two nonempty open sets). –  Syd Henderson Sep 9 '13 at 21:44
    
@Syd: Yes, I am; unfortunately, I forgot to mention that by convention in that paper all spaces are Tikhonov (and non-discrete, though that doesn’t matter here) unless otherwise stated. –  Brian M. Scott Sep 9 '13 at 21:48

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