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I was reading a proof of the following theorem in my textbook:

A set $A$ is closed iff $A' \subseteq A$.

Proof: Suppose $A$ is closed and $x \in A'$. If $x \notin A$, then $x\in A^c$, an open set. Thus $\mathcal{N}(x, \delta)\subseteq A^c$ for some positive $\delta$. But then $\mathcal{N}(x, \delta)$ can contain no points of $A$. Thus $x$ is not an accumulation point of $A$ and so $x\notin A'$, a contradiction. We conclude that $x\in A$. Therefore $A'\subseteq A$.

Now suppose $A'\subseteq A$. To show that $A$ is closed, we show $A^c$ is open. If $A^c$ is not open, there is $x\in A^c$ that is not an interior point of $A^c$. Therefore, no $\delta$-neighborhood of $x$ is a subset of $A^c$; that is, each $\delta$-neighborhood of $x$ contains a point of $A$. This point must be different from $x$, since $x\in A^c$. Thus $x\in A'$. But $A'\subseteq A$, so $x\in A$. This is a contradiction. We conclude that $A$ is closed.

Somewhere along the proof, the complement of $A$ is proven to be open. In order to achieve this, the author first assumes that it's not open and produces a contradiction. But isn't that wrong since it's possible for a set to be neither open nor closed?

Community
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cppcoder
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    What exactly is wrong? The proof starts with the assumption that $A^c$ is not open; it is not assuming that $A^c$ is closed. – angryavian Jan 08 '19 at 22:43
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    To show that $P,Q$ are equivalent, you can prove $P \implies Q$ and $\lnot P \implies \lnot Q$. – copper.hat Jan 08 '19 at 22:47
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    He makes absolutely no assumptions about whether $A^c$ is closed. He only assume $A^c$ is not open. He gets a contradiction which means..... It is false that $A^c$ is not open. That means $A^c$ is open. That's all. – fleablood Jan 08 '19 at 22:48
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    As one of my professors used to say: “sets are not doors.” – Jim Jan 08 '19 at 23:11

2 Answers2

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No problems here. Note that the author doesn't assume $A^c$ to be closed. Just that it isn't open. And since "open" means "for any point in the set, we have a certain property", we get that "not open" means "for at least one point in the set, we don't have that property". The point $x$ they choose is one such point guaranteed to exist by that assumption.

Arthur
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There are four possiblities

1) $A^c$ is open and closed.

2) $A^c$ is open and not closed.

3) $A^c$ is not open and closed.

4) $A^c$ is not open and not closed.

He assumed: $A^c$ is not open. That means either $3$ or $4$. He got a contradiction. That shows that both $3$ and $4$ are both false. Thus the only possibilities are $1$ or $2$. So he concludes either $1$ or $2$ is true. Which means.... $A^c$ is open.

However as he didn't give a carnivore's fig whether $A^c$ was closed or not he made no mention of it at all.

So it comes down to:

I) $A^c$ is open. or

II) $A^c$ is not open.

He assumes II) gets a contradiction, and concludes I). $A^c$ is open. Now maybe $A^c$ is closed. Or maybe it isn't. He doesn't care and made no assumptions to the fact.

But he does know, and correctly so. That $A^c$ is open....

fleablood
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