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Consider the following statements:

${\bf A.}$ $\exists $ $s \in \mathbb{R}$ and $\exists $ integer $N>0$ such that for all $\varepsilon > 0$ and all $n>N$ one has $|s_n - s| < \varepsilon $

${\bf B.}$ $\forall $ $s \in \mathbb{R}$ and $\exists \varepsilon > 0$ such that for all $n \in \mathbb{N}$ one has: $|s_n - s| < \varepsilon $

${\bf C.}$ $\forall $ $s \in \mathbb{R}$ and $\exists \varepsilon > 0$ and some $n \in \mathbb{N}$ one has: $|s_n - s| < \varepsilon $

Now, $A$, $B$ and $C$ may look similar, but I want to understand their differences:

What I think:

${\bf A}$ We have that $s$ and $N$ are fixed at the beginning. And any $\epsilon > 0$ and after an index $n$ then the diference $|s_n-s|$ increases without bound. Does this mean that the sequence is ${\bf unbounded}$?

${\bf B}$ Let $s$ be given. Now, $\epsilon > 0$ is a function of $s$. Isnt condition $B$ true for every sequence?

${\bf C.}$ This one looks very similar to $B$, but here only one term of the sequence, say $N$ satisfies $|s_N - s|<\epsilon $. Isn't this also satisfied trivially by any sequence?

Are my interpretations correct? Can someone help with mastering the quantifiers which can be a little difficult...

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  • $\begingroup$ For (A) you're thinking of the role of $\epsilon>0$ in the wrong way. What happens if $\epsilon$ is very small? Note that in words (the end of) the sentence is saying that $|s_n-s|$ is smaller than every positive number. What is the only number $\geq 0$ with this property? $\endgroup$
    – peek-a-boo
    May 23 '20 at 8:21
  • $\begingroup$ it is 0! so A is equivalent to convergence of sequence? $\endgroup$ May 23 '20 at 8:23
  • $\begingroup$ No, it is not equivalent to convergence of a sequence, but it definitely implies convergence. (But convergence of a sequence doesn't imply (A)). I'm currently in the proceess of writing up the correct answers, but I'll leave it to you to fill in the details and justify the claims I made. $\endgroup$
    – peek-a-boo
    May 23 '20 at 8:24
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Let's analyze them one at a time.

First for $A$, your interpretation is incorrect. Statement $A$ implies that there is a number $s>0$ and a number $N \in \Bbb{N}$, such that for all $n > N$, we have $s_n = s$. In other words, the sequence is eventually constant. (As a consequence, you can deduce that the sequence is convergent with limit equal to $s$, and hence bounded). But note that the converse is clearly false: not every convergent sequence is eventually constant.

For $B$, you're right that $s\in\Bbb{R}$ is given first and then $\epsilon>0$ depends on the given $s$. However, no, statement $B$ is not always true, because it actually implies that the sequence $\{s_n\}$ is bounded (and as you know, not every sequence is bounded). Try to prove that the following two statements are equivalent:

($B$) For every $s \in \Bbb{R}$, there is an $\epsilon > 0$ such that for all $n \in \Bbb{N}$, one has $|s_n - s| < \epsilon$.

($B'$) There is an $s \in \Bbb{R}$, and there is an $\epsilon > 0$ such that for all $n \in \Bbb{N}$, one has $|s_n - s| < \epsilon$. (this is usually the definition of boundedness of a sequence... perhaps with different notation)

In general of course, you can't change a "for all" to "there exists", but in this case, they turn out to be equivalent.

Finally, for $C$, you're right that it is trivially satisfied by every sequence. Why? Because given any $s \in \Bbb{R}$, I simply CHOOSE (i.e I'm telling you the existence part) $n=1$ and $\epsilon = |s_1-s| + 1$. Then, clearly $|s_1 - s| < \epsilon = |s_1 - s| + 1$.

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  • $\begingroup$ I think $B' $ and $B$ are not equivalent. Only $B \implies B'$ since if true for all $s$, then it is true for just one $s$, but I am afraid the other direction may not be true $\endgroup$ May 23 '20 at 21:29
  • $\begingroup$ @Theoneandonly Like I mentioned in the answer, in general you can't change a "for all" to "there exists". But in this very specific situation, the two statements are equivalent. $B \implies B'$ is trivially true for the reason you mentioned. The harder direction of $B \Leftarrow B'$ is what I leave to you to prove. (Hint: triangle inequality is helpful here) $\endgroup$
    – peek-a-boo
    May 23 '20 at 21:53

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