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I am working on a pretty straight forward proof, I am trying to show that when I have a family of topological spaces $ (X_i, T_i )_{i \in I}$ where all $(X_i, T_i )$ are connected that the product topology is then also connected. I kinda know how to prove this but I think I am having some sort of misunderstanding, because say I am looking at the product topology produced by two topological spaces, would $(X_1\times \emptyset) \cup (\emptyset \times X_2)$ not be the disjoint union of two open sets in the product topology?

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  • $\begingroup$ That set is empty, so it’s certainly not the disjoint union of two non-empty open sets. $\endgroup$ – Brian M. Scott Jun 24 '16 at 18:58
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\begin{align*} X_1\times\varnothing\neq&\,X_1\text{, nor }X_1\times X_2;\\ X_1\times\varnothing=&\,\varnothing. \end{align*}

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  • $\begingroup$ Hmm okay I did not know that the set would be considered empty. But would $(X_1×∅)∪(∅×X_2)$ not still be $X_1×X_2$? $\endgroup$ – lelman Jun 24 '16 at 19:03
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    $\begingroup$ $\varnothing\times X_2=\varnothing$, too. The union of two empty sets is still empty... $\endgroup$ – triple_sec Jun 24 '16 at 19:06
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    $\begingroup$ @lelman: No: that set is $\varnothing\cup\varnothing=\varnothing$. Any product $A\times\varnothing$ or $\varnothing\times A$ is empty: there is no possible ordered pair in it. $\endgroup$ – Brian M. Scott Jun 24 '16 at 19:07
  • $\begingroup$ Ah okay now I understand my mistake, I thought of the union as $(X_1∪∅)×(∅×X_2)$ which is obviously incorrect, thanks for the help! $\endgroup$ – lelman Jun 24 '16 at 19:12
  • $\begingroup$ For that, let $x\in A,y\in B$ with $A,B$ any sets. Then by definition $x,y\notin \varnothing$ thus $(x,y)\notin \varnothing\times X$ for any set $X$. $\endgroup$ – David Molano Jun 24 '16 at 19:41

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