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Let $A$ and $B$ be disjoint closed subsets of $\mathbb{R}^n.$ How can we show that if $X\setminus A$ and $X\setminus B$ are path-connected, then $X\setminus (A\cup B)$ is path-connected?

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    $\begingroup$ What is $X$ here? $\endgroup$ – Nate Eldredge Mar 9 '13 at 16:32
  • $\begingroup$ Sorry for that mistake, here as X It was meant $\mathhbb{R}^n$ $\endgroup$ – Shohruh Mar 11 '13 at 14:42
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If you let me use the hammer called "Mayer-Vietoris sequence", you have an exact sequence $$ \rm H^1(\mathbb R^n) \to H^0(\mathbb R^n \backslash(A \coprod B)) \to H^0(\mathbb R^n \backslash A) \oplus H^0(\mathbb R^n \backslash B) \to H^0(\mathbb R^n) \to 0$$

But the first term is zero, and $\rm H^0(\mathbb R^n \backslash A) = H^0(\mathbb R^n \backslash B) = H^0(\mathbb R^n) = \mathbb Z$ because they are all connected. So $ \rm H^0(\mathbb R^n \backslash(A \coprod B)) = \mathbb Z$ which means it is connected, hence path-connected because $\mathbb R^n$ is locally path-connected.

I am looking for a simpler proof, but at least this proof tells you that the answer relies heavily on the fact that $\rm H^1(\mathbb R^n) = 0$, so this theorem is wrong for a random path-connected space $\rm X$.

For example, take $\rm X$ be the circle, $\rm A$ and $\rm B$ two disjoint closed arcs. Then $\rm X \backslash (A \cup B)$ has two components.

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