If I have two compact subsets, $A$ and $B$, of the plane $\mathbb{C}$, and we know that $\partial A$ and $\partial B$ are homeomorphic, can we say that $A$ and $B$ are homeomorphic?


Let $A$ be a full circle and $B := ∂A$.

  • $\begingroup$ Is $\partial$ idempotent in general? $\endgroup$ – Kenny Lau Sep 15 '17 at 8:51
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    $\begingroup$ @KennyLau, no, consider $\mathbb Q \subseteq \mathbb R$, but it is for closed sets. $\endgroup$ – Mees de Vries Sep 15 '17 at 8:53

No. Consider a closed annulus, such as $\{z \in \mathbb C \mid 1 \leq |z| \leq 2\}$, and the union of two closed discs, such as $\{z \in \mathbb C \mid \min(|z-2|,|z+2|) \leq 1\}$.

  • $\begingroup$ In this case, $A$ and $B$ are homeomorphic because $A$ is connected while $B$ isn't. $\endgroup$ – Kenny Lau Sep 15 '17 at 8:49
  • $\begingroup$ how about if $A$ and $B$ are both connected, and $A$, $B$ have non-empty interior? $\endgroup$ – Yee Neil Sep 15 '17 at 10:32
  • $\begingroup$ Still no; consider a "pinched annulus", where the two boundary circles touch at one point, and the non-disjoint union of two circles $\{z \in \mathbb C \mid \min(|z-1|,|z+1|) \leq 1\}$ $\endgroup$ – Mees de Vries Sep 15 '17 at 10:36
  • $\begingroup$ Ok, I got it ! Maybe I need more conditions on this question! $\endgroup$ – Yee Neil Sep 15 '17 at 11:32

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