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Suppose that $\text{int}(A) \subseteq B \subseteq A$, is it true that $B \cup \text{bdy}(A) = A$?

Note: $\text{bdy}(A) $ is the boundary of $A$

Consider $A = [-1,1]$, the $\text{int}(A) = (-1,1)$, and $B = [-1,1)$, then $B \cup \text{bdy}(A) = A$.

Does this property generalize to all sets in Euclidean space?

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    $\begingroup$ If you change $A$ to $[-1,1)$ in your example, you'll see that $B\cup\mathop{\rm bdy}(A)$ strictly contains $A$ but also contains $1$. $\endgroup$ – Greg Martin Sep 11 '19 at 2:04
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We know that $ A^0\cup \partial A = \overline{A} $ ($ A^0 $ is the interior, $ \partial A $ is the boundary, and $ \overline{A} $ is the closure of $ A $). It is true that $ B\cup \partial A \supset A $ but not necessarily true that it equals $ A $. For instance, let $ A = (0,1) $ and let $ B=A $. Then $ B\cup \partial A = [0,1] $ which strictly contains $ A $.

We'd only have equality if $ A $ was closed (where $ A = \overline{A} $).

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    $\begingroup$ Beautiful. So it is true that $B \cup \text{boundary}(A)$ is always equal to the closure of $A$. $\endgroup$ – Cauchy's Carrot Sep 11 '19 at 2:07
  • $\begingroup$ Yup, this is true since $A^0\subset B\subset A $. $\endgroup$ – poopstraw Sep 11 '19 at 2:09
  • $\begingroup$ @Cauchy'sCarrot No, see my example where that fails too. $\endgroup$ – Henno Brandsma Sep 11 '19 at 4:44

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