All convergent sequences are bounded confusion We proved that all convergent sequences are bounded. However, when proving the following:
If $x_n$ converges to $x$ and $y_n$ converges to $y$, then $\dfrac{x_n}{y_n}$ converges to $\dfrac{x}{y}$,
we use the fact that if a sequence $y_n$ converges to $y$, with $y_n, \ y \not = 0$, then there exists a positive real number $c$ and a natural number $N$ such that $|y_n| \geq c$ for $n> N$.
My question is, since $y_n$ is convergent, then it should be bounded seeing as every convergent sequence is bounded. But, doesn't $|y_n| > c$ imply that it $y_n$ is not bounded?
 A: You need to take care of your quantifiers. In fact, you need to start by quantifying what you write, to give it meaning. Writing "$\color{red}{y_n>M}$" carries no meaning by itself. 
Unbounded (from above) is 

"For every $M>0$ there exists $n$ such that $\color{red}{y_n>M}$.

What you have is 

"There exists $M>0$ such that, for every $n$, $\color{red}{y_n>M}$. 

See the difference?
A: I hope I'm not misunderstanding your question.
That there exists a positive real, $c$, which is smaller than $|y_n|$ for $n > N$ does not imply that there does not exist a different positive real, $C$, which is greater than $|y_n|$ for all $n$.
A: Consider an example. $x_n = 1$ and $y_n = 1+\frac{1}{n}$ So $x_n \rightarrow 1$ and $y_n \rightarrow 1 \neq 0$. 
Let us take $c = \frac{1}{2}$. Then $y_n > c$ $\forall$ $n \in \mathbb{N}$.
It does not imply that $\{y_n\}$ is unbounded. For being unbounded above we have to show that for any  $G > 0 $ there is an $N$ depending on $G$ s.t. $y_N > G$. It does not hold here.
