"If there exists $Y {\subsetneq} X$ such that $P$ holds, then there exists $x {\not\in} Y$ such that $Q$ holds"

Question

Let $X$ be a nonempty set.

If there exists a subset $Y \subsetneq X$ such that property $P$ holds, then there exists an $x \not \in Y$ such that property $Q$ holds.

I'm trying to determine the contrapositive of the above conditional statement. My initial guess:

• If for every $x \not\in Y$, $\neg Q$ holds, then for every $Y \subsetneq X$, $\neg P$ holds.

But this guess's hypothesis hasn't introduced $Y$ or said anything about what $Y$ is!

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EDIT

This is the exercise that I was trying to take the contrapositive of:

Let $f : M \to N$ be a map between $R$ modules. Suppose that for some maximal ideal $\mathfrak m \subsetneq R$, the induced map $M / \mathfrak m M \to N / \mathfrak m N$ satisfies some property $P$. Prove that there exists $r \not\in \mathfrak m$ such that $R_r \otimes M \to R_r \otimes N$ satisfies some property $Q.$

Answer 1

Let $X$ be a nonempty set.

If there exists a subset $Y \subsetneq X$ such that property $P$ holds, then there exists a $x \not \in Y$ such that property $Q$ holds.

Has the set $Y$ been specified or pre-defined?

1. If not, then the above string $$\big(\exists Y {\subsetneq} X\, P(Y)\big)\implies\exists x {\not\in} Y\, Q(x)\tag0$$ isn't even a statement (it's technically an open formula), as variable $Y$'s second occurrence is free, so you have probably misread the original text, and the original statement is likely this instead: $$\forall Y {\subsetneq} X\,\Big( P(Y)\implies\exists x {\not\in} Y\, Q(x)\Big).\tag1$$
2. If yes, then the original statement is more helpfully (less confusingly) written as $$\big(\exists \boldsymbol S {\subsetneq} X\, P(\boldsymbol S)\big)\implies\exists x {\not\in} Y\, Q(x)\tag2$$ instead.

Now there should no longer be a problem taking contrapositive.

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Reply to the OP's comment

Ah that's helpful. I suppose it's the way the problem is written? It goes:

Let $f : M \to N$ be a map between $R$ modules. Suppose that for some maximal ideal $\mathfrak m \subsetneq R$, the induced map $M / \mathfrak m M \to N / \mathfrak m N$ satisfies some property $P$. Prove that there exists $r \not\in \mathfrak m$ such that $R_r \otimes M \to R_r \otimes N$ satisfies some property $Q.$

This is what I was trying to take the contrapositive of.

Ah, then your source text indeed means sentence $(1)$ rather than formula $(0)$ or sentence $(2):$ $$\forall Y {\subsetneq} R\,\Big( A(Y)\implies\exists r {\not\in} Y\, B(r)\Big).$$ Your mistake is understandably misunderstanding the instruction ‘suppose that for some maximal ideal $m\subsetneq R\text’$ as ‘if there exists a maximal ideal $m\subsetneq R\text’$ rather than ‘if for an arbitrary maximal ideal $m\subsetneq R\text’;$ after all, this $m$ is even referred to again in the conclusion/consequent. (Formalising mathematical induction has the same pitfall, explained here.) Then the contrapositive is just $$\forall Y {\subsetneq} R\,\Big( \forall r {\not\in} Y\, \lnot B(r)\implies \lnot A(Y)\Big).$$