Understanding some variations of the limit definition of a sequence

Question

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...

Answer 1

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 $|s_1 - s| < \epsilon = |s_1 - s| + 1$.