# Understanding some variations of the limit definition of a sequence

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...

• 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? May 23 '20 at 8:21
• it is 0! so A is equivalent to convergence of sequence? May 23 '20 at 8:23
• 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. May 23 '20 at 8:24

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

• 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 May 23 '20 at 21:29
• @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) May 23 '20 at 21:53