Negation of a conjuntion in a actual proof. 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
 A: Suppose you have proved that (0) if $A$ and $B$ then contradiction!
Then all these five follow from your conclusion (obviously, I hope!)


*

*It isn't the case that $A$ and $B$ [or if you prefer that in symbols, $\neg(A \land B)$]

*Either not-$A$ or not-$B$ or maybe both [$\neg A \lor \neg B$]

*If $A$ is true, then $B$ isn't. [$A \to \neg B$]

*If $B$ is true, then $A$ isn't. [$B \to \neg A$]

*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.]
