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Let $A$ be a connected subspace of $X$ and suppose $A\subseteq Y\subseteq\overline{A}$. Prove that $Y$ is connected.

My attempt: Suppose that $Y$ is not connected. Then $Y=U_1\cup U_2$ where $U_1$ and $U_2$ are non-empty, disjoint open sets in $X$. Now WLOG what if $A$ sits entirely in $U_1$? This cannot happen because $Y\subseteq\overline{A}$, so for every $y\in Y$, every open set containing $y$ intersects $A$. So in particular, $U_1$ and $U_2$ intersect $A$, and $(A\cap U_1)\cap (A\cap U_2)=\varnothing$ with $A=(A\cap U_1)\cup (A\cap U_2)$. But $(A\cap U_1)$ and $(A\cap U_2)$ are also both non-empty and open. So this contradicts our hypothesis that $A$ is compact.

Does anyone see a problem with my argument?

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    $\begingroup$ swap compact with connected and that looks good to me. $\endgroup$ – Dan Rust Mar 21 '14 at 7:13
  • $\begingroup$ "So this contradicts our hypothesis that $A$ is compact".. really? $\endgroup$ – user87543 Mar 21 '14 at 7:13
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    $\begingroup$ Yeah, except for the typo "A is compact" (I'm sure you meant connected) it looks good $\endgroup$ – chriseur Mar 21 '14 at 7:14
  • $\begingroup$ Oops. Yeah, I meant A is connected :) $\endgroup$ – user124910 Mar 21 '14 at 7:28
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Suppose that $Y$ is not connected. Then $Y=U_1\cup U_2$ where $U_1$ and $U_2$ are non-empty, disjoint open sets in $X$.

They should be open in $Y$ since you cannot assume that $Y$ is open in $X$.

Now WLOG what if $A$ sits entirely in $U_1$? This cannot happen because $Y\subseteq\overline{A}$, so for every $y\in Y$, every open set containing $y$ intersects $A$. So in particular, $U_1$ and $U_2$ intersect $A$,

If $U_1$ doesn't intersect $A$, there must be a $y$ in $(Y-A)\cap U_1$ and then $U_1$ intersects $A$ since $y\in\partial A$. Similarly for $U_2$. So both intersect $A$. I think that's your reasoning here.

and $(A\cap U_1)\cap (A\cap U_2)=\varnothing$ with $A=(A\cap U_1)\cup (A\cap U_2)$. But $(A\cap U_1)$ and $(A\cap U_2)$ are also both non-empty and open.

Open in $A$ :-)

So this contradicts our hypothesis that $A$ is compact.

Connected ;-)

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Use the following very useful facts:

A topological space is connected if and only if every continuous map from it into a discrete space is constant.

If $f$ is continuous then $f(\overline A) \subset \overline{f(A)}$.

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because $A$ is connected so $\bar{A}$ is allso connected, by definition of $\partial A$ we cant separate any element of $\partial A$ from the interior of $A$ by tow open sets, and because $Y$ is the union of the interior of $A$ and elements of $\partial A$ so $Y$ is connected

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