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Let $(X,\tau)$ be a topological space and $E,Y\subset X$ such that $E$ is connected, $E\cap Y\neq\emptyset$ and $E\cap(X\setminus Y) \neq \emptyset$. Thus $E\cap\partial Y \neq \emptyset$. Any help is appreciated. Thanks.

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closed as off-topic by user21820, user99914, GNUSupporter 8964民主女神 地下教會, Shailesh, DRF May 18 '18 at 14:09

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Recall that $X$ is the disjoint union of $Y^\circ$, $\partial Y$, and $(X\setminus Y)^\circ$. Suppose $E\cap\partial Y=\varnothing$, Then $E$ is the disjoint union of $E\cap Y^\circ$ and $E\cap(X\setminus Y)^\circ$. But this is a separation of $E$ (in the subspace topology), a contradiction. It follows that $E\cap \partial Y\ne\varnothing$.

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    $\begingroup$ @user1337 The proper way to thank someone on this site is to accept the answer (the check mark). You can also upvote it (the up arrow); you can upvote more than one answer. For future questions - they will be better received if you include what you tried and where you are stuck rather than just asking for a proof. $\endgroup$ – Ethan Bolker May 18 '18 at 13:45

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