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Let $A$ and $B$ are compact subsets of $\mathbb{C}$ such that $B=A \cup$ iso($B$). If $A^c$ is connected then prove that

$B^c$ is connected.

Here iso($B$) denotes the set of isolated points of $B$.

For a compact subset $A$ of $\mathbb{C}$ we say that $A^c$ is connected if $A^c$ does not have any bounded component.

Manu Rohilla
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Since $A^c$ is an open connected set of $\mathbb C$, it is path connected. The set of isolated points is always countable in $\mathbb R^n$[Proof: $\mathbb R^n$ is second countable. Subspaces of second countable spaces are countable. An uncountable set consisting of isolated points only is not second countable.] Now $B^c$ is equal to $A^c$ minus countably many points. Can you conclude that $B^c$ is path connected using Why is open connected minus countable set is connected?

YuiTo Cheng
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  • How can we say that if $B$ is compact then isolated points of $B$ are finite. – Manu Rohilla Apr 18 '19 at 13:48
  • @ManuRohilla I deleted my previous comment (since it is very wrong! E.g. ${\frac{1}{n}}_{n\in \mathbb N}$ is an infinite set consisting of isolated points only) and make a significant edit, please check out. – YuiTo Cheng Apr 18 '19 at 14:06