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I proved that if $Y$ is a proper subspace of a Banach space then interior of $Y$ is empty. But, looking at what I logically did, made me confused.

To me it looks like I'm proving that $$(A \wedge B) \Longrightarrow \neg A,$$ which implies $(A \wedge B)$ is false, hence its negation is true, i.e $(\neg A \vee \neg B)$.

But why am I in postion to choose the implication $$A \Longrightarrow \neg B?$$ I.e If a subspace is closed then it has empty interior. And does it matter how I used the interior definition in arriving at the contradiction?

The proof is the same as in this answer Every proper subspace of a normed vector space has empty interior

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user123124
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  • I'm not sure if I understand your question completely (or at all), but it might help to see that "if $A$ then not $B$" is precisely the negation of "$A$ and $B$". – Dániel G. Sep 21 '15 at 16:48

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Suppose you have proved that (0) if $A$ and $B$ then contradiction!

Then all these five follow from your conclusion (obviously, I hope!)

  1. It isn't the case that $A$ and $B$ [or if you prefer that in symbols, $\neg(A \land B)$]
  2. Either not-$A$ or not-$B$ or maybe both [$\neg A \lor \neg B$]
  3. If $A$ is true, then $B$ isn't. [$A \to \neg B$]
  4. If $B$ is true, then $A$ isn't. [$B \to \neg A$]
  5. Either $A \land \neg B$ or $\neg A \land B$ or $\neg A \land \neg B$.

Indeed these are all equivalent. There's nothing to chose (though one way of putting things might be more immediately natural, depending on the context).

In particular, you might want (as in the present context) to stress that if $A$ is true, then $B$ isn't true, i.e. $A \to \neg B$. That's fine: that follows from (and indeed is equivalent to) what you've proved.

Of course, what you can't do, what would be illegitimate, is to pick just one of the three disjuncts in (5) and conclude that. But that isn't what you are doing in asserting (3), is it?

[If you are confused about basic logical points like this, Daniel Velleman's How to Prove It (CUP) is a much recommended resource, and will well repay a couple of afternoon's study.]

Peter Smith
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  • Thats where I get stuck, I dont see how choosing in 5 is different from 3 – user123124 Sep 21 '15 at 17:09
  • Well, if the conditional is truth-functional, (3) follows from each disjunct in (5). So you don't have to choose one disjunct to conclude (3) from (5). In the other direction, (5) follows from (3). But just from the information that (3) is true you can't infer which disjunct of (5) is true. – Peter Smith Sep 21 '15 at 17:20
  • The crucial point, however, is that that (3) evidently follows from (0). You don't need to go through (5); but if you do you don't have to choose one of the disjuncts to proceed. – Peter Smith Sep 21 '15 at 17:27
  • Hi again! Something struck me today. When doing a proof by contradiction we basically take $A$ to be the negation of what we wanna prove then use some fact(or truth) $B$ about the objects involved and derive a contradiction by being clever! e.g https://math.stackexchange.com/questions/451700/prove-that-sqrt-5-is-irrational $A={\sqrt{5}$ is rational} and $B$=${$any rational is quotient of some $p,q}$ – user123124 Feb 04 '21 at 20:28