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Let $X_i, \; i\in I$ be connected topological spaces and let $X=\Pi_{i\in I}X_i.$

I am familiar with the theorems that $X$ is a connected space in the product topology but not in the box topology. However, I've started to prove the opposite of the claim in the box topology case and it brings me to a result that $X$ must be connected, which of course is wrong, but I cannot find where am I missing the point. Here is my attempt:

Suppose $X$ is not a connected space, i.e., there exist a separation $X=A\cup B.$ The canonical projections $\pi_i:X\to X_i$ are continuous open surjections in the box topology, hence $X_i=\pi_i(A)\cup\pi_i(B)$ is a separation of $X_i$ which is a contradiction. Thus X is connected in the box topology.

Does this proof work in the case of product topology?

Emo
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1 Answers1

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The proof is incorrect for normal products too (as well as for box products).

Your claim that $\pi_i[A]$ and $\pi_i[B]$ are disjoint is nonsense. Projections are very non-injective in general and only for injective (1-1) functions $f$ we can safely say that $A,B$ disjoint implies that $f[A]$ and $f[B]$ are as well. Draw some counterexamples in the plane $\Bbb R^2$ (disjoint lines will often have full projections onto both factors) to see how badly this fails.

For the box topology there is a sort of anti-connectedness theorem: any infinite family of $X_i$,$i \in I$ where all $X_i$ are $T_{3}$ and have at least two points have a non-connected box topology product. (so even $\Bbb R^{\Bbb N}$ in the box topology is disconnected e.g. ) A proof can be found in the handbook of Set-Theoretic Topology.

Henno Brandsma
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