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I was thinking about a counter example of a set that isn't a linear continuum because it doesn't satisfy the least upper bound property with the dictionary topology, but I can't found anyone.

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  • $\begingroup$ $\mathbb Q$? $ $ $\endgroup$ Dec 12 '19 at 15:01
  • $\begingroup$ Well yes, but I'm searching for an counter example in $\mathbb{R^2}$ $\endgroup$
    – user732763
    Dec 12 '19 at 15:04
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    $\begingroup$ $\mathbb Q \times \mathbb Q$? $\endgroup$ Dec 12 '19 at 15:13
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As I noted recently, $[0,1] \times [0,1)$ is not a linear continuum as $\{0\} \times [0,1)$ has an upper bound but no lub.

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