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Prove that the cantor space is totally disconnected.

Let $(G,T)$ be the Cantor space and let $\prod_{i=1}^{\infty}(A_i,T_i)$ be homeomorphic to the Cantor space where $(A_i,T_i) = (\{0,2\}, T_{discrete})$.

I know that a product space is totally disconnected iff each of its factors are totally disconnected and totally disconnectedness is preserved through homeomorphisms. Therefore $(G,T)$ is totally disconnected iff $(\{0,2\}, T_{discrete})$ is totally disconnected.

But $(\{0,2\}, T_{discrete})$ is not totally disconnected because the components $C_{\{0,2\}}(0)$ and $C_{\{0,2\}}(2) \neq $ a singleton set.

So wouldn't this imply that the Cantor space is not totally disconnected?

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    $\begingroup$ if you cant see that $\{0,2\}$ is disconnected, read your definitions again... $\endgroup$ – Forever Mozart Jun 24 '16 at 17:20
  • $\begingroup$ Totally disconnected set - Every non-empty connected subset is a singleton set. Or $C_{X}(a) = \{a\}$ for each $a \in X$. Connected set - The only subsets that are clopen are the null set and the set itself. If $\{0,2\}$ was totally disconnected, then $C_{\{0,2\}}(0) $ = a singleton set. But since it's discrete, $\{0,2\}$ contains $2$ and $\{2\}$ contains $2$ and their union = $\{0,2\}$. What am I misunderstanding? $\endgroup$ – Oliver G Jun 24 '16 at 17:28
  • $\begingroup$ $C_{\{0,2\}}(0)\neq C_{\{0,2\}}(2)$. The next-to-last sentence in your comment isn't relevant to the definition. $\endgroup$ – Eric Stucky Jun 24 '16 at 17:30
  • $\begingroup$ I re-worded it for clarity. $\endgroup$ – Oliver G Jun 24 '16 at 17:31
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    $\begingroup$ I see now, $\{0,2\}$ is not connected because $\{0\}$ is clopen and $\{2\}$ is clopen, so $\{0\}$ and $\{2\}$ are connected subsets because they're both the only connected set that is a singleton set. $\endgroup$ – Oliver G Jun 24 '16 at 17:40

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