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My book has two adjacent exercises, in which $\partial A$ is the set of boundary points of $A$ (a subset of some metric space $X$) and $\overline{A}$ is the closure of $A$. The first asks me to prove that $\overline{A} = A \cup \partial A$, which was simple and is discussed in this question.

The second exercise asks me to prove that $\partial A = \overline{A} \cap \overline{A^c}$. I proved this one way that matches the book's solution, but I also have another proof that uses the first exercise, the fact that $\partial A = \partial A^c$, and DeMorgan's Laws.

\begin{align*} \overline{A} \cap \overline{A^c} &= \left(A \cup \partial A\right) \cap \left(A^c \cup \partial A^c\right) \\ &= \left(A \cup \partial A\right) \cap \left(A^c \cup \partial A\right) \\ &= \partial A \cup \left(A \cap A^c\right) \\ &= \partial A \cup \varnothing \\ &= \partial A \end{align*}

Is this a correct proof?

Community
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2 Answers2

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For the second,

$\bar{{A}}$ $\cap$ $A^\mathrm{o}$ = ($A^\mathrm{o}\cup\partial{A}$) $\cap$($(A^c)^\mathrm{o}\cup\partial{A^c}$) = $\bar{{A}} \cap ((A)^\mathrm{o})^c$ = $\partial A$

PrimusSanguis
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I don't quite see how you go from $(A \cup \partial A) \cap (A^c \cup \partial A)$ to the line below it in one step.

I would expand to 4 terms: $(A \cap A^c) \cup (\partial A \cap A^c) \cup (A \cap \partial A) \cup (\partial A \cap \partial A)$, where the first term equals $\emptyset$, the final one $\partial A$ and the middle terms $(\partial A \cap A) \cup (\partial A \cap A^c)$ can be regrouped as $\partial A \cap (A \cup A^c) = \partial A \cap X = \partial A$ as well, so the total group becomes indeed $\partial A$.

Of course the definition of the boundary already gives the formula almost directly: $x \in \partial A$ iff every open neighbourhood of $x$ intersects $A$ (i.e. $x \in \overline{A})$ and every open neighbourhood of $x$ intersects $X \setminus A = A^c$ (i.e. $x \in \overline{A^c}$. So $\partial A = \overline{A} \cap \overline{A^c}$, almost by definitions of boundary and closure.

Henno Brandsma
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