# Why is this equality relevant in this proof of closedness?

I've got a theorem:For any set $A$ in a metric space: $\left(\overline{A}\right)'=A'$

and the book proves a corollary: $\overline{A}$ is closed.

The proof is this: $\left(\overline{A}\right)'=A'\subset \overline{A}$, meaning $\overline{A}$ is closed.

I think it's pretty obvious from the theorem but why they added $A'\subset \overline{A}$ to the proof? is there something I'm missing?

They’re using the fact that a set $S$ is closed if and only if it contains all of its limit points, i.e., if and only if $S'\subseteq S$. Thus, they can show that $\operatorname{cl}A$ is closed by showing that $(\operatorname{cl}A)'\subseteq\operatorname{cl}A$, and this is exactly what they’ve done.

• Yes, but what is the role of $A'$ in here? Jul 3 '16 at 0:00
• @José: They’re proving that $\operatorname{cl}A$ is closed as a corollary of the result that $(\operatorname{cl}A)'=A'$. Because they’ve already proved that, then know that $(\operatorname{cl}A)'=A'$, and they also know that $A'\subseteq\operatorname{cl}A$. Putting the two together, they deduce that $(\operatorname{cl}A)'=\operatorname{cl}A$, which is exactly what they needed in order to conclude that $\operatorname{cl}A$ is closed. Jul 3 '16 at 0:03
• AH I get it now -________- now I feel stupid Jul 3 '16 at 0:21
• @José: Don’t beat yourself up too much: it happens to all of us from time to time. Jul 3 '16 at 0:29

I believe one reason people find math tricky is the shorthand proofs that textbooks favor which still require a lot of thought to fill in hidden assumptions and definitions. The proof is more obvious written this way...

A′⊂ cl A by definition of closure. Using the above theorem we replace A′ so (cl A)′ ⊂ cl A.

Therefore cl A is closed as it contains all its limit points.