Logic statements equivalence

Question

Are the following two statements logically equivalent? Or does the second imply the first? Please explain.

(1) $\forall x\in X$ $\exists$ $y\in$ Y: $P(x,y)$

(2) $x\in X$ $\iff$ $\exists y\in Y$ :$P(x,y)$

I'm trying to generalize the following concepts from a logic point of view

Let $(X, \tau)$ be a topological space.

Suppose for a subset A of X , $int(A)=A$ Then $x \in A$ $\iff$ there exists an open set U containing x so that x $U\subset A$

Which my professor said implies

For each $x\in A$ there exists an open set U containing x so that x $U\subset A$

It completely make sense, but I'm trying to see and understand if it may not be the case for (1) and (2) above

Answer 1

$\forall x\in X\, (S)$ is an abbreviation for $\forall x\,(x\in X\implies S).$

Similarly, $\exists y\in Y\,(T)$ is an abbreviation for $\exists y\,(y\in Y\land T).$

In your Q, statement (2) is not a "sentence" because $x$ occurs as a free variable. We can fix this by preceding it with "$\forall X$".( Much mathematical writing has "implicit" or "understood" $\forall$'s, as it can be tedious to always put them in). Now we have $$(1)\iff \forall x\,(x\in X\implies \exists y\,(y\in Y\land P(x,y)\,)).$$ $$ (2)\iff \;\forall x\,(x\in X\iff \exists y\,(y\in Y\land P(x,y)\,)).$$ Note that (1), unlike (2), allows that there might exist some $x\not \in X$ and $y\in Y$ such that $P(x,y).$

Answer 2

The second is stronger than the first because $P(x,y)$ could hold with $x\notin X$. This is possible with (1), not with (2).

Answer 3

No, they are not equivalent.

Consider the integers as domain of discourse, and let $P(x,y)$ be the property that $2y=x$. Then if we let $X=\{2,4,6\}$ and $Y=\mathbb Z$ it is easy to see that the first formula is satisfied for this $X$ and $Y$, but the second one isn't (e.g. take $y=5$, then $P(10,5)$ holds, but $10\notin X$).

The equivalence you search for is probably the following:

• $\forall x\in X(\varphi(x,\bar v))$
• $\forall x(x\in X\to\varphi(x,\bar v))$

This is equivalent by definition, since it is how we define the abbreviation "$\forall x\in X$". Of course from this we can prove that $\forall x(x\in X\leftrightarrow\varphi(x,\bar v))$ implies the second sentence.

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As for your topology example: since the second sentence you mention implies the first, your professor can indeed say this.

The reverse statement is that if every $x\in A$ is contained in an open $U\subset A$, then $x\in A$ if and only if there exists an open $U\subset A$ such that $x\in U$.

This is not true because of the rules of first-order logic, since we need to use what the meaning of $\subset$ is to be able to prove it. The $\Leftarrow$ direction of the if and only if is proved by the definition of $\subset$ as "$U\subset A$ iff $x\in U$ implies $x\in A$".