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

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?

• 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 $1$. – Greg Martin Sep 11 '19 at 2:04

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 $$A = \overline{A}$$).
• Beautiful. So it is true that $B \cup \text{boundary}(A)$ is always equal to the closure of $A$. – Cauchy's Carrot Sep 11 '19 at 2:07
• Yup, this is true since $A^0\subset B\subset A$. – poopstraw Sep 11 '19 at 2:09