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Does there exist $A \subseteq \mathbb{R}^{2}$ with usual topology and $A$ not enumerable such that $\tau_{A}$ the subspace topology in $A$ it is a discrete space?

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HINT: $\Bbb R^2$ has a countable base $\mathscr{B}$. If $\langle A,\tau_A\rangle$ is a discrete space, then for each $x\in A$ there is a $B_x\in\mathscr{B}$ such that $B_x\cap A=\{x\}$. The map $A\to\mathscr{B}:x\mapsto B_x$ is injective (one-to-one), so ... ?

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  • $\begingroup$ @Brain: Your answer is also nice; +1) $\endgroup$ – Paul Jun 5 '13 at 3:37
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No. Note that $A$ is second countable and metrizable. Then $A$ must be separable. This is a contradiction with $A$ is discrete and uncountable!

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No. If it were so then we could create a collection of open subsets of the plane, one for each point of the space, all of which are disjoint. But each open subset would contain at least one point with rational coordinates which is not in any other such open subset. Thus, the cardinality must be less than or equal to that of the rationale, making it enumerable.

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    $\begingroup$ This is fairly seriously incomplete. Relative discreteness of $A$ directly tells you only that there are open sets $V_x$ for $x\in A$ such that $V_x\cap A=\{x\}$ for each $x\in A$; it takes moderately significant work to show that they can be chosen to be pairwise disjoint. In particular, it’s entirely possible to have a separable Tikhonov space with an uncountable discrete subset — even an uncountable closed discrete subset. $\endgroup$ – Brian M. Scott Jun 5 '13 at 3:25
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    $\begingroup$ But in a metric space, can't you just take a ball with radius the minimum distance to the nearest point, then halve it (for each point)? $\endgroup$ – Brian Rushton Jun 5 '13 at 3:29
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    $\begingroup$ Yes, you can, but I think that it’s a significant enough step to require some explicit acknowledgement, if only in the form of a Why? appended to the statement. $\endgroup$ – Brian M. Scott Jun 5 '13 at 3:32
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    $\begingroup$ This is the second time today that my intuition about discrete sets has been screwy... thanks for the example! $\endgroup$ – Brian Rushton Jun 5 '13 at 3:33
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    $\begingroup$ @Paul: The new version looks fine; +1. $\endgroup$ – Brian M. Scott Jun 5 '13 at 3:35

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