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I am trying to prove that (connected) components of a topological space are connected. I'll first define what I mean by a 'component of a topological space':

For a topological space $X$, write $x\sim y$ if $\exists\ Y \subset X$ such that $Y$ is connected and $x, y \in Y$ (this is an equivalence relation. The components of $X$ are the equivalence classes for $\sim$.

My proof so far of why components are connected:

Let $C$ be a component, $x_0 \in C$. Then $\forall y \in C, \exists A_y \subset X$ such that $A_y$ is connected, with $x_0, y \in A_y$. Note that $A_y \subset C, C = \bigcup_{y \in C}A_y$...

Note sure how to finish this off. Any insight would be appreciated!

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  • $\begingroup$ Hint:Lemma: If $Y_1$ and $Y_2$ are connected subspaces of $X$ and $Y_1\cap Y_2\ne \phi$ then $Y_1\cup Y_2$ is connected. $\endgroup$ – DanielWainfleet May 24 '16 at 14:30
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HINT: Suppose that $C$ is not connected. Then there are open sets $U$ and $V$ such that $U\cap C$ and $V\cap C$ are disjoint and non-empty. Say $x_0\in U\cap C$, and pick $y\in V\cap C$. Now use $U$ and $V$ to show that $A_y$ cannot be connected after all.

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  • $\begingroup$ Cheers; to clarify, your assumptions lead to the fact that $A_y$ cannot contain both $x_0$ and $y$, and this is the contradiction. Is this correct? $\endgroup$ – Joe May 24 '16 at 10:05
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    $\begingroup$ @Joseph: Since $A_y$ was defined to contain those points, I'd put it a little differently and say that the contradiction is that the set $A_y$ must be both connected and not connected. $\endgroup$ – Brian M. Scott May 24 '16 at 10:09

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