Closure is the Smallest Closed Set

Theorem

Let $A \subseteq R$ . Then $\overset{―}{A}$ (the closure of $A$ ) is the smallest closed set containing $A$ .

The proof isn't super trivial, since we might accidentally throw in too many points.

Proof

If $L$ is the set of limit points of $A$ , then it is immediately clear that $\overset{―}{A}$ contains the limit points of $A$ . There is still something more to prove, that because taking the union of $L$ with $A$ then we could potentially produce some new limit points of $\overset{―}{A}$ . See 32 Open and Closed Sets Practice#3.2.7 for practice on this.

Now any closed set containing $A$ must contain $L$ as well. This shows that $\overset{―}{A} = A \cup L$ is the smallest closed set containing $L$ .

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