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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?

Nick
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Warz
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3 Answers3

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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?

Pedro
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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 $n$.

Nick
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  • Yes that was my question, I just thought it did imply it. I'll have another think. – Warz Jan 07 '14 at 16:16
  • You should consider Hopeless Fool's recent answer--it provides a concrete counterexample to that implication. I'd chance to say you're getting confused with the difference between "there exists" and "for all." – Nick Jan 07 '14 at 16:21
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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.

Supriyo
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    Yes we can. Pick $G=.1$. You need to find an $N$ for ALL integers (or positive reals) $G$. –  Jan 07 '14 at 17:22
  • You have written things quite incorrectly. Unboundedness (from above) is stated as follows: "For every $M>0$ there exists an $n$ (depending on $M$) such that $a_n>M$." That is, for each $M$ we can find an element of the sequence that beats it. Certainly we don't need "for all $n>N$". $0,1,0,2,0,3,0,4,\ldots$ is unbounded, but doesn't satisfy a condition over all its tails. – Pedro Jan 07 '14 at 17:26
  • Check if it is correct now. – Supriyo Jan 08 '14 at 01:40