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If $X$ is a connected topological space, and $Y$ is a connected subspace, then for any connected component $C$ of $X\backslash Y$, $Y \cup C$ is connected.

Results proven before: Connected components partition the topological space, are themselves connected and closed.

Useful Example for visualization: $X=\mathbb{R}^2$, $Y=\mathbb{R} \times \{0\}$ and $C = \mathbb{R} \cup (0, \infty)$

Attempt 1: Suppose on the contrary that $Y \cup C$ isn't connected, let $Y \cup C = A \cup B \cap (Y \cup C)$, for $A$ and $B$ open sets of $X$ with empty intersection and non-trivial. Without loss of generality, since $Y$ is connected we may assume that $Y=A\cap (Y \cup C)$,$C=B\cap (Y \cup C)$. This shows that $C$ is a closed subset of $X\backslash Y$ and an open subset of $Y \cup C$, therefore $X\backslash(Y \cup C) = U \cap (X\backslash Y)$ and $C = V \cap (Y \cup C)$, applying the union we find $X\backslash(Y \cup C) \cup C = X\backslash Y = U \cup (V \cap C)$ Therefore: $$ X = X\backslash Y \cup Y = U \cup (V \cap C) \cup (A \cap (Y \cup C))$$ This union is disjoint but not by open sets, and I can't seem to reduce it to open sets in any way.

David Melo
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This is false. Here is a counterexample. Define

$$A = \{ 1/n \mid n \in \mathbb N \},$$ $$Y = (0,1] \times\{0\},$$ $$B = Y \cup A \times [0,1] ,$$ $$C = \{(0,1)\} ,$$ $$X = B \cup C .$$

$B$ is connected. Thus its closure $\overline B = B \cup (\{0\} \times [0,1])$ in $\mathbb R^2$ is connected and since $B \subset X \subset \overline B$, also $X$ is connected.

$Y$ is a closed connected subspace of $X$.

We have $X \setminus Y = A \times (0,1] \cup C$. The components of $X \setminus Y$ are the sets $\{1/n\} \times (0,1]$ and $C$. But clearly $Y \cup C$ is not connected.

Remark.

Under additional assumptions on $Y$ and $X \setminus Y$ we get a correct theorem. See Connected and locally Connected Space. Show $\overline{U}\cap C \neq \emptyset$ for C close and U connected component in $X\setminus C$.

Paul Frost
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