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I'm triying to construct a bounded subset $A$ of $\mathbb{R}^2$ such that $int(A)=\emptyset$ but $int\left(\overline{A}\right)\ne \emptyset$ I can't see how to start, i've tried to use $A=\{(1/n,0):n\in\mathbb{N}\}$ it's interior is empty but also the clousure's interior. Any hint? Or generic example? I'm using the usual topology of the plane.

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  • $\begingroup$ I know that example of rationals could work, but I don't know if the intersection with some bounded "ball" makes them work $\endgroup$ – Ragnar1204 Feb 22 '17 at 2:15
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    $\begingroup$ Try the set of all points in the unit square with rational coordinates. $\endgroup$ – user49640 Feb 22 '17 at 2:16
  • $\begingroup$ I'll try right now thanks! $\endgroup$ – Ragnar1204 Feb 22 '17 at 2:17
  • $\begingroup$ "I know that example of rationals could work, but I don't know if the intersection with some bounded "ball" makes them work" it would. It absolutely would. So any bounded set in R^2 intersected Ted with Q^2 ( or even just QxR) will do. $\endgroup$ – fleablood Feb 22 '17 at 3:24
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In $\mathbb R$, one can consider $A:=\mathbb Q \cap [0,1]$, which has empty interior, but whose closure is all of $[0,1]$. Using this same example, we consider

$$A \times A \subseteq \mathbb R^2.$$

Since the usual topology is equivalent to the product topology (boxes), we know that the closure of $A \times A$ is the unit square, which has nonempty interior.

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