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I need help with this problem.

If $A\subset X$, and $\partial A$ and $X$ are connected, then $Cl(A)$ is connected.

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    $\partial A$ is allowed to be empty? No assumptions on $A$? – 2017-01-08
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    @HennoBrandsma Seems like it. My proof doesn't need any more assumptions, at least. Remember that empty sets are connected. If $\partial A$ is empty, then $A$ is both open and closed in $X$, and since $X$ is connected, that means $A = X$ or $A = \emptyset$, so $Cl(A) = X$ or $\emptyset$ is connected. It works out either way. – 2017-01-08
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    Why is the question tagged (convex-analysis)? – 2017-01-08

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Hint: Prove the contrapositive. In other words, assume that $X$ is connected but $Cl(A)$ is not, and prove that $\partial A$ must be disconnected.

If $Cl(A)$ is disconnected, then there are $A_1, A_2 \subset Cl(A)$ that are non-empty and closed (in $Cl(A)$) such that $A_1\cap A_2 = \emptyset, A_1\cup A_2 = Cl(A)$. Since $A_1, A_2$ are closed in $Cl(A)$, which is closed in $X$, we have that $A_1, A_2$ are closed, non-empty and disjoint in $X$. This means that they cannot be open in $X$. Therefore, $\partial A_1$ and $\partial A_2$ are both non-empty, and disjoint, and closed in $X$, and therefore in $A$ and in $\partial A$. But $\partial A_1\cup \partial A_2 = \partial A$, which means that $\partial A$ is disconnected.

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    Yes, I thought about doing that, but I do not know how. @arthur – 2017-01-08
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    @lilian I added more details. – 2017-01-08
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    Ok, thanks por help! @arthur – 2017-01-08
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    You could reference http://math.stackexchange.com/questions/218805/the-boundary-of-union-is-the-union-of-boundaries-when-the-sets-have-disjoint-clo as a result you're using. It's not widely taught. – 2017-01-08
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    @HennoBrandsma I could. But why stop there? I could write a whole book proving every single little thing I need to get to this result, starting from the definition of a topology. Or I could write a single paragraph, trusting that anyone who reads it is capable of filling in the details as they go. – 2017-01-08