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I read somewhere that a connected set in the digital line topology is a subset of consecutive integers. Recall that the digital line topology is a topology on $\mathbb{Z}$ with basis elements $\{n\}$ for $n$ odd and $\{n-1, n, n+1\}$ for $n$ even. However, if we consider the set $\{2,3,5\}$, then this set must be disconnected based on the above statement, as it is missing the integer $4$. In order to show that this set is disconnected, we must show that it has a separation (or that it is the union of two disjoint nonempty open sets). $\{5\}$ is open in the digital line topology, as $5$ is odd. However, for the case of the set $\{2,3\}$ it's trickier. We can show that this set is not open, as we cannot construct an open neighborhood around $\{2\}$ such that the neighborhood is entirely contained in the two-tuple set. How do I show that this set is indeed disconnected?

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The set $\{2,3\}$ is open, since it is equal to $\{1,2,3\}\cap\{2,3,5\}$ and $\{1,2,3\}$ is open set in $\mathbb Z$.

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  • $\begingroup$ Such a simple oversight. Thank you. $\endgroup$ – user312437 Nov 28 '18 at 22:58
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To say that a subset is disconnected is to say that it is disconnected under the subspace topology. To show that $\{2,3,5\}$ is disconnected, you can show that it is a disjoint union of sets that are open under the subspace topology. Remember that a set $S$ is open under the subspace topology here if $S=T\cap\{2,3,5\}$ where $T$ is open in the original topology, in this case, the digital line topology. Then $\{2,3,5\}=\{2,3\}\cup \{5\}$, the former of which is open because $\{2,3\}=\{1,2,3\}\cap\{2,3,5\}$, and $\{1,2,3\}$ is open in the digital line topology, and the latter is open as you pointed out.

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