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For a subset $A$ of topological space $X$, the exterior of $A$ is the set $\mathrm{ext}\,(A)=X\setminus\mathrm{cl}(A)$.

What I have so far:

$\text{ext}\,(A) = X\setminus \text{cl}(A) = X\setminus (A \cup A') = (X\setminus A) \cap (X\setminus A') $ (DeMorgan's Laws)

But this is where I am stuck. I am unsure if this proof is a simple transformation, or if i'm approaching it from the wrong angle.

Any advice?

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  • $\begingroup$ Can you see that $\text{cl}(A) = X \setminus \text{int}(X\setminus A)$? $\endgroup$ – user417289 Nov 7 '17 at 21:28
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Since$\DeclareMathOperator{\Int}{Int}$ $A \subseteq \overline{A}$, we have $$X\setminus \overline{A} \subseteq X\setminus A$$ Hence, $X\setminus \overline{A}$ is an open set contained in $X\setminus A$, so it is also contained in its interior. Therefore, $X\setminus \overline{A} \subseteq \Int(X\setminus A)$.

Similarly, from $\Int(X\setminus A) \subseteq X\setminus A$ we obtain $$A = X \setminus (X \setminus A) \subseteq X \setminus \Int(X\setminus A)$$ Therefore, $X \setminus \Int(X\setminus A)$ is a closed set which contains $A$, so it also contains its closure:

$$\overline{A} \subseteq X\setminus \Int(X\setminus A)$$

By complementing, we obtain $\Int(X\setminus A)\subseteq X\setminus \overline{A}$.

Therefore:

$$\mathrm{Ext}\,A = X\setminus \overline{A} = \Int(X\setminus A)$$

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Whenever you’re asked to prove $A=B$, for two sets $A$ and $B$, you assume $x \in A$ and show $x \in B$. Then you assume $x \in B$ and show $x \in A$

If $x \in ext(A)$, then $x$ is in an open set that does not intersect $A$. Hence $x$ is not in the closure of $A$.

If $x \in X$ and also $x$ is not in the closure of $A$ then it is not a limit point of $A$. Hence there is an open set containing $x$ that does not intersect $A$. Hence it’s in the exterior of $A$.

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