Difference b/w - $ \exists x \in \mathbb{R} \forall y \in \mathbb{R} x+y>0; \space \forall x \in \mathbb{R} \exists y \in \mathbb{R} \space x+y>0$? Consider the following statements:
(a) $ \exists \space x \in \mathbb{R} \space \forall \space y \in \mathbb{R} \space x+y>0$
(b)$\forall \space x \in \mathbb{R} \space \exists \space y \in \mathbb{R} \space x+y>0$
Why is the statement (a) different from (b)? Why can't they become equivalent on simply interchanging the variables? Are we considering $y$ as a function of $x$? 
The answer to (a) is given as
(a) is false. Since its negation $∀x ∈ \mathbb{R} \space ∃ \space y ∈ \mathbb{R} \space x + y ≤ 0$ is true. Because if $x ∈ \mathbb{R}$, there
exists $y ∈ \mathbb{R}$ such that x + y ≤ 0. That is, we get a counterexample, $y = −(x + 1)$ which gives $x + y = x − x −1 = −1 ≤ 0$.
Here for all $y$, did we mean for all function of $x$?
The answer to (b) is given as: (b) is true.
Thanks in advance...
 A: It is convention that variables bind sequentially.  Hence, in (a), variable $x$ must be chosen BEFORE looking at $y$; hence a single $x$ must work for all $y$.  
In (b), again $x$ is chosen before $y$, which in this case works in our favor.  We choose $y$ after choosing $x$, so the $y$ can vary depending on which $x$ we have.  Different $x$'s will lead to different $y$'s, since we only need one $y$ (not all of them).  So long as we can find such a custom $y$ for every possible $x$, (b) is true.  For example, we can use $y=1-x$.
A: The first one says that
there exists an x such that
for all y,
x+y > 0.
In other words,
x > -y for all y.
But if y is large and negative
(y = -|x|-1, for example),
this is false.
The second says that
for each x there is a y
such that
x+y > 0.
In other words,
there is a y such that
y > -x.
This is true,
as can be seen
by choosing
y = |x|+1.
A: The first case will be false as you said earlier and the reason for this is that y is a dependent variable as opposed to x. If you set a single value x, it is not true for all y such that $x+y >0$. Here is a generalized counter example. Let x = x, and if I take $y=-x -1$, then $x+y=-1$ and -1 is less than $0$ which make it false, so it is not true. 
Where as in the second case you can set any y in term of all x to make the inequality true for you. There is a flexibility in the second case, as it is absent in the first case.
Hope it will help you.
A: They aren't just switching the variables.
The first says "there is some $x$ such that all $y$ interact with that $x$ in a certain way.
The second says "for every x, there is some $y$ that interacts with that instance of $x$ in a certain way."
An analogy might be:
(a): There exists a man such that all men are his father.
(b): For every man, there exists a man who is his father.
It is clear here that, as in your example, (a) is false, (b) is true, and the statements mean completely different things.
A: "Every person has a surname".
"There is a surname that every person has." 

Why is the statement (a) different from (b)? Why can't they become equivalent on simply interchanging the variables? Are we considering y  as a function of x ? 

Kind of. We say the quantifier binding $y$ is nested inside the scope of the quantifier binding $x$.
The order of binding matters when the quantifiers differ; it produces statements with quite different meaning.
Let's consider a tally table.
$\forall x\in \text{Rows}\;\exists y\in \text{Columns} : \text{Checked}(x,y)$ - says every row has a check in at least one column.  These may be different columns and so columns need not have checks in every row.
$\exists y\in \text{Columns}\;\forall x\in \text{Rows} : \text{Checked}(x,y)$ - says there is a column which has a check in every row.
A simpler example would be $\forall x \exists y : y=x$ versus $\exists y \forall x : x=y$.  That is "everything is equal to something" versus "something is equal to everything".
