Logical Statements - Proofs

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Let $f$ be a real-valued function (a function with target space the set of reals). Let $P(x, M)$ stand for $|f(x)| \leq M $, let $N$ be the set of positive real numbers, and let $\mathbb{R}$ be the set of real numbers.

a) Which of the following statements is an accurate translation of "f is bounded"?

(i): ($\forall M \in N$)($\exists x \in \mathbb{R}$)($P(x,M)$)

(ii): ($\exists M \in N$)($\forall x \in \mathbb{R}$)($P(x,M)$)

(iii): ($\forall x \in \mathbb{R}$)($\exists M \in N$)($P(x,M)$)

(iv): ($\exists x \in \mathbb{R}$)($\forall M \in N$)($P(x,M)$)

I understand that (III) is the answer that defines a bounded function, but I don't understand how it differs from (II). Also, if someone can provide me with a more explicitly method of reading these types of statements that would really help clarify a lot of things.

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(ii) Exists an $M$ such that for each $x$ real, $|f(x)| \leq M$.

(iii) For each $x$ real there exists an $M$ such that $|f(x)| \leq M$.

In the first case $M$ is unique for all $x$.

In the second case $M$ depends on $x$. In this sense, we can restate (iii) as follows:

(iii) For each $x$ real there exists an $M(x)$ such that $|f(x)| \leq M(x)$.

The function is bounded if there is an $M$ such that for all $x$ you have $|f(x)| \leq M$. Then the solution is.... (ii)!

Example

Consider $f(x) = x^2+1$ which is clearly unbounded.

(ii) is not satisfied, while (iii) is satisfied. Indeed, for all $x$, take for example $M(x) = x^2+2$. Then:

$$|f(x)| = |x^2 + 1| = x^2 + 1 \leq x^2 +2 = M(x)$$

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The difference between : $\exists x \forall y$ and $\forall y \exists x$ is clearly shown by this example regarding natural numbers :

$\forall n \ \exists m \ (n < m)$

is clearly true in $\mathbb N$,

while :

$\exists m \ \forall n \ (n < m)$

is false in $\mathbb N$.