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Let $(X, \mathcal{T})$ be a topological space, and $X = U \cup V$ such that $U, V \neq \emptyset$, $U \cap V = \emptyset$, for open $U, V$. Let $C$ be path-connected. I want to prove that $C$ must be a subset of either $U$ or $V$.

Attempt (I did not get very far): suppose $U \cap C \neq \emptyset$ but $C$ is not a subset of $U$. Then $C \backslash U \neq \emptyset$. If $C \backslash U$ and $C$ are open, then $C = (C \backslash U) \cup (C \cap U)$ which are disjoint non-empty open sets which contradicts the connectedness of $C$, but I'm not convinced that is true, and I don't know any other way to prove this (or if this is even true, because it's just a lemma I need to prove that the deleted comb space is connected.

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    $\begingroup$ Is $C$ path connected? $\endgroup$ Dec 19, 2017 at 19:18
  • $\begingroup$ @AndresMejia yes $\endgroup$ Dec 19, 2017 at 19:23
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    $\begingroup$ You are correct, but you need to mention that path connectedness implies connectedness. $\endgroup$ Dec 19, 2017 at 19:23

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Assume for a contradiction that $C \cap U$ and $C \cap V$ are both nonempty.

Let $x \in U \cap C$ and $y \in V \cap C$ . Since $C$ is path connected, there is a map $\alpha:[0,1] \to X$ so that $\alpha(0)=x$ and $\alpha(1)=y$ But then take the image of $\alpha$ and intersect it with $U,V$ respectively, under the subspace topology, the components are open, so this constitutes a separation, meaning that the image is disconnected. This is a contradiction since $[0,1]$ was connected.

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  • $\begingroup$ Thanks. Just out of curiosity: does the statement still hold if $C$ is connected but not path-connected? $\endgroup$ Dec 19, 2017 at 19:24
  • $\begingroup$ Yes. This is similarly straightforward. The intersection with $U,V$ constitutes a separation under the subspace topology $\endgroup$ Dec 19, 2017 at 19:25

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