Let $(X,d)$ be a metric space and let $A \subset X$.

Prove that:

(a): $X-int(A)=\overline{X-A}$

(b): $X-\overline{A}=int(X-A)$

(Notation: $\overline{E}$ is the closure of $E$ where $\overline{E}=E\cup E'$, $E'$ is the set of all limit points of $E$.)

I'm assuming that we will use the concept that a set $S$ is closed iff $S^c$ is open.

Also I know that the closure of a set is always closed.

Other than that, I don't know how I should approach this. Any advice will be greatly appreciated. Thanks in advance!


1 Answer 1


For the first one,

Let $x \in X - int(A)$. Then by definition of interior, we have that every ball around $x$ intersects $A^{c}$. Hence $x \in \overline{X-A}$. The reverse inclusion is proven by reversing the above steps.

For the second one,

let $x \in X - \overline{A}$. Then $x \notin A$ and $x \notin A^{\prime}$. Thus, there exists a ball around $x$ that does not intersect $A$, which means that $x \in int(X-A)$. Again the converse is proven by reversing the steps.

These proofs are decidedly terse (as you asked for hints), and if you need clarification, feel free to ask.


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