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I have come across the proof that the Cartesian product of totally disconnected sets is also disconnected in the following post.

Product of totally disconnected space is totally disconnected?

  1. However does the above theorem extends to arbitrary product spaces also? (I.e. A component where $x_\alpha$ and $y_\alpha$ does not differ can

  2. Also, what would happen if we have two distinct points such that the component for x is always a subset of y? (I.e. having $x_\alpha= \{\frac{1}{3}\}$ and $y_\alpha= \{\frac{1}{3},\frac{2}{3}\}$, will the proof still works?

  3. Finally is it possible to have X to be totally disconnected but having some $X_i$ to not be totally disconnected?

Thank you!

Community
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user90831
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1 Answers1

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For (3): If some $X_i$ is not totally disconneted, there is a connected component $Y_i \subseteq X_i$ which is not a singleton. Now choose $x_j \in X_j$ for every $j \in I \setminus \{i\}$, and define
$$ \hat Y_i := Y_i \times \prod_{j \in I\setminus\{i\}} \{x_j\} $$ Then $\hat Y_i$ is homeomorphic to $Y_i$ and hence connected and not a singleton. But $\hat Y_i$ is a subspace of $X$. Contradiction.

martini
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