Show that the boundary of a set equals the boundary of its complement

Solution 1:

If you know that $\mathrm{closure}(A)=(\mathrm{int}(A^c))^c$, then you also have $\mathrm{closure}(A^c)=(\mathrm{int}(A))^c$ because $(A^c)^c=A$. Therefore

$$\mathrm{boundary}(A)=\mathrm{closure}(A)\cap(\mathrm{int}(A))^c = \mathrm{closure}(A)\cap \mathrm{closure}(A^c).$$

The last expression is symmetric in $A$ and $A^c$.

Solution 2:

A point in X is said to be a boundary point of a set A if each neighborhood of the point intersects both A and X\A. If you replace A with the complement of A in the statement, you get the same statement. So very simply both the sets have the same boundary.