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I am trying to show the following two statements:

1) Let $U$ be a connected open subset of $\mathbb{C}^n$ and let $L$ be a closed subset of $U$. Show that $$accum(U\setminus L)\cap L\neq \emptyset,$$ where accum() denotes the set of accumulation points.

2) Let $L$ be a closed subset of $\mathbb{C}^n$ and let $x\in L$. Show that there exists an open connected neighborhood $U$ of $x$ in $\mathbb{C}^n$ such that $U\setminus L$ is connected and $$x\in accum(U\setminus L).$$

Edit: For (1) $L$ is strictly contained in $U$.

For (2) $x\in \partial L$ and the complement of $L$ in $\mathbb{C}^n$ is connected.

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  • $\begingroup$ The second claim seems wrong. What if $L$ is a closed ball, and $x$ is its center? Are you maybe supposed to assume that $x\in\partial L$? $\endgroup$ – G Tony Jacobs May 30 '17 at 17:29
  • $\begingroup$ 1 is false. U = L = C^n is counterexample. $\endgroup$ – William Elliot May 31 '17 at 2:17
  • $\begingroup$ @GTonyJacobs yes I am sorry, I suppose $x\in \partial L$. Can we show it if so? $\endgroup$ – Joe May 31 '17 at 8:58
  • $\begingroup$ @WilliamElliot for 1 I suppose $L$ is strictly contained in $U$. If so there is still a counter example? $\endgroup$ – Joe May 31 '17 at 9:01
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1) This holds for each connected topological space $U$ and its non-empty closed subset $L\ne U$. Indeed, put $V=U\setminus L$. Clearly, $accum V\subset \overline{V}$. On the other hand, $accum V\supset \overline{V}\setminus V$. Thus

$$(accum V)\cap L=(accum V) \cap (U\setminus V)= (accum V)\setminus V=\overline{V}\setminus V.$$

Thus if $(accum V)\cap L=\emptyset$ then the set $V$ is closed and $U=L\cup V$ is a partition of a connected space into two its disjoint proper subsets, a contradiction.

2) If $x\in \partial L$ and the complement of $L$ in $\mathbb{C}^n$ is connected then we can put $U=\Bbb C^n$ :-) and $V=U\setminus L$. Indeed, then $$x\in (\partial L)\cap L=(\partial (U\setminus L)) \cap L= (\partial (U\setminus L)) \setminus (U\setminus L)=(\partial V)\setminus V\subset accum V.$$

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  • $\begingroup$ Thank you very much! For the second statement, does the same conclusion hold if we replace $\mathbb{C}^n$ by a connected open subset $C\subset \mathbb{C}^n$ with $L$ having connected complement in $\mathbb{C}^n$ not in $C$? $\endgroup$ – Joe Jun 2 '17 at 11:06

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