# $\forall x \in I , \exists y \in I$ such that $xy \in I$

I just have a small question! Really basic I'm sure but something is bothering me. Take note of the following statement:

$\forall x \in I , \exists y \in I$ such that $xy \in I$

Does this statement say that for all $x$ in $I$, there exists a $y$ in $I$ or does it say that for every single $x$ in $I$, there exists a $y$ that makes the statement true.

So for example if we took $x1$ to be $\pi$ and $x2$ to be $\sqrt 2$, can we assign them a unique $y$ for each one or for both of them?

It is a very fine difference but in proving the statement, it is rather confusing. I understand that the statement is true however but would like to solidify my reasoning.

Thank you :)

Ah yes, I understand now! Thank you everyone :) Really appreciated!

• The statement says "For all x in I there exists a y, that is in I, such that the multiple xy is also in I." It does mean ANY x, as long as x is in I. But remember, y also has to be contained in I for this statement to hold true. Apr 11, 2013 at 2:38
• The $y$ depends of the value of $x$, that is, for $x_1$ you have a $y_1$ and for $x_2$ you have a $y_2$. Apr 11, 2013 at 2:40

It means both: for all $x \in I$ there exists a $y \in I$...is equivalent to "for every single x in I, there exists a $y$ that makes the statement true.

But what this means, with respect to your follow up questions, is that each $x$ can have some particular y (which may depend on that x) for which the statement is true. It does not mean that there must be one y that satisfies the statement for every x. So for any particular x, there can exist a particular y, such that the statement is true. As $x$ varies, so can the "some particular" y.

So for example,

$$\forall x \in \mathbb R,\;\exists y \in \mathbb R\;( x \lt y)$$ is true. For any $x$ there is some y that is greater than that $x$. If $x = 1/2$, then there exists a $y = 1$ suffices to make $x\lt y$ true. If $x = 4$, then there is some $y$, say $5$, for which $x\lt y$.

This is different that $$\exists y \in \mathbb R,\;\forall x\in \mathbb R \;(x < y)$$ which is false: there is no $y \in \mathbb R$ that is greater than every $x \in \mathbb R$.

You might appreciate this related post, which helps clarify the difference between the $\exists y \forall x$... and $\forall x \exists y$...You'll also find a list of posts linked to that post, which address the same question.
In many contexts, $\forall x\in\mathcal{I},\,\exists y\in\mathcal{I},\,...$ means given an $x$ there is a $y\in\mathcal{I}$ (that may depend on that given $x$) such that...
If you want to say "for every single $x$ in $\mathcal{I}$, one "fixed" $y$ works fine," then you would write $\exists y\in\mathcal{I},\,\forall x\in\mathcal{I},\,..$ Note the difference between the two expressions.
Therefore, $\forall x\in\mathbb{Z},\,\exists y\in\mathbb{Z},\,x+y=1$ is true, as we can choose $y = 1-x$. On the other hand, $\exists y\in\mathbb{Z},\,\forall x\in\mathbb{Z},\,x+y = 1$ is false: Assume such a $y$ exists. Then, pick $x = -y+123\in\mathbb{Z}$, then $x+y = 123$.
Edit: To make things more readable, I sometimes write, for example, $\forall x\in\mathcal{I},\,\exists y(x)\in\mathcal{I},\,...$ instead of $\forall x\in\mathcal{I},\,\exists y\in\mathcal{I},\,...$ in order not to get confused. Perhaps this may also work for you.