# Proving $0$ is partial limit of $a_n$ - Why such an $n$ exists s.t. it satisfies $a_n \leq 0$ , $a_{n+1} >0$?

Problem: If sequence $$(a_n)$$ has 10,-10 as partial limits and in addition $$\forall n \in \mathbb{N}.|a_{n+1} - a_n| \leq \frac{1}{n}$$, then 0 is a partial limit of $$(a_n)$$.

Answer (Proof): We need to show $$\forall \epsilon > 0 \forall N \in \mathbb{N} \exists n \geq N . | a_n| < \epsilon$$.
Let $$\epsilon >0$$ and $$N \in \mathbb{N}$$ be arbitrary. We know $$\frac{1}{n} \rightarrow 0$$, Hence there exists $$N_1 \in \mathbb{N}$$ s.t. $$\forall n>N_1 . \frac{1}{n} < \epsilon$$ . Since -10 is a partial limit then there exists $$n_1 \geq max \{ N, N_1 \}$$ s.t. $$-1 < a_{n_1} - ( -10 ) < 1$$ , specifically $$a_{n_1} < -9$$ . Since 10 is also a partial limit, there exists $$n_2 > n_1$$ s.t. $$-1 < a_{n_2} - 10 < 1$$, specifically $$a_{n_2} > 9$$. Since $$a_{n_1} < 0$$ and $$a_{n_2} > 0$$ and also $$n_2 > n_1$$, $$\color{red}{ \text{there exists} }$$ $$n_1 \leq n < n_2$$ $$\color{red}{ \text{s.t.} }$$ $$a_n \leq 0$$ , $$a_{n+1} > 0$$. In addition $$n \leq N_1$$, hence $$a_{n+1} < a_{n+1} - a_{n} = | a_{n+1} - a_{n} | \leq \frac{1}{n} < \epsilon$$ and $$n+1 \geq N$$, as we wanted.

My Difficulty: I understood everything up to the line "$$\color{red}{ \text{there exists} }$$ $$n_1 \leq n < n_2$$ $$\color{red}{ \text{s.t.} }$$ $$a_n \leq 0$$ , $$a_{n+1} >0$$. " I don't understand why such an $$n$$ exists s.t. it satisfies $$a_n \leq 0$$ , $$a_{n+1} >0$$. Do we get this $$n$$ from some kind of instantiation here? ( Maybe from instantiation in the definition for the partial limits 10 & -10 ? )
Related: Prove $0$ is a partial limit of $a_n$ - In this linked thread first answer is virtually the same proof above here but my question remains the same because I have still don't understand from there why such an $$n$$ exists.

• Do you know what it means to say $\mathbb{N}$ is well-ordered? Commented Apr 23, 2021 at 18:45
• @BrianMoehring I haven't learned about it so no, but it'd be nice to have that as an additional explanation since I'll learn about it in the future. Commented Apr 23, 2021 at 18:47

## 2 Answers

You have an infinite number of values near -10 and another infinite number of values near 10. So pick one value near -10, pick a higher point in the sequence near 10. Since the step sizes are under $$\frac 1 n$$ between each term, the step sizes are smaller than 1, so the only way to get from the negative number (say at step 100) to the positive number (say at step 200) is by crossing 0 at some step in between. (For instance, it may occur at step 152 as the last negative before we go positive). Basically it's a variant concept of the intermediate value theorem.

We then get an infinite number of those terms, each of whom are within $$\frac 1 n$$ of $$0$$, so we get a third subsequence getting as close to 0 as you want.

Using your concept (And dropping the primes) first pick $$\epsilon>0$$. For simplicity, assume $$\epsilon<1$$, if you picked a bigger value it'll work for 1 also.

Now we know there's some $$n_1'$$ such that $$|{a_n}_1-10|<\epsilon$$ Now, using that as our new $$N$$, we know there is an $$n_2$$ such that $$n_2>n_1$$ and $$|{a_n}_2+10|<\epsilon$$

So since the first value is within 1 of 10 and the second is within 1 of -10, by transitivity we get that $$a_{n_2}\leq -9\leq 9\leq a_{n_1}$$

Now, for simplicity, pick out a monotone decreasing subsequence within the values between $$n_1$$ and $$n_2$$, you can do so by just tossing out every value that is not less than the previous term. So now I have some monotone sequence of numbers $$m_1=n_1$$ through $$m_r=n_2$$ such that for $$\forall s \text { s.t} 1\leq s Since we have $${a_m}_s-{a_m}_{s+1}<\frac 1 {m_s}$$, the step size is always less than one. Since we are travelling a distance from greater than 9 to less than -9 monotonically in steps less than size 1, it takes us at least 18+ steps to get there. Since they are monotone, for some particular $$s'$$, $$a_s'>0$$,$$a_{s+1}'<0$$. These two points are within $$\frac 1 {s'}$$ of each other and are on either side of 0, so we get $$|a_s'|<\frac 1 {s'}$$.

That generated our first value near 0. Repeat ad infinitim to generate new terms closer and closer to 0, as the related $$s'$$ you find will all be higher numbers

• I think it'll be easier for me if I'd see the derivation of $n$ in terms of instantiation ( maybe we even use Archimedean property here ), I know that: Since 10 is partial limit then: $\forall \epsilon>0 \forall N \in \mathbb{N} . \exists n'_1>N. | a_{n'_1} - 10 | < \epsilon$. Since -10 is partial limit then: $\forall \epsilon>0 \forall N \in \mathbb{N} . \exists n'_2>N. | a_{n'_2} + 10 | < \epsilon$. How do I instantiate these statements to get $n$ ? Commented Apr 23, 2021 at 18:35
• editing answer for this
– Alan
Commented Apr 23, 2021 at 18:40
• There's an explicit construction for you. Also, on a side note, this was fun. I've been stuck teaching two semesters worth of basic math and intermediate algebra, nice to actually do a touch of formal math again :) @hazelnut_116
– Alan
Commented Apr 23, 2021 at 18:58
• I tried the following, but I think it is wrong, do you have any idea why?: Suppose for all $n_1 \leq n < n_2$ that $a_n < 0$ . Then $a_{n_2-1} < 0$ ( note that indeed $n_1 \leq n_2 -1 < n_2$ ) , then $| a_{n_2} - a_{n_2 - 1} | > 9$ ( since $a_{n_2} > 9$ and from the last assumption we have $a_{n_2 -1 } < 0$ , note that this inequality has nothing to do with the given fact that $| a_{n+1} - a_n | < \frac{1}{n}$ for all $n$ ). Now since we know that $| a_{n+1} - a_n | < \frac{1}{n}$ for all $n$, Commented Apr 23, 2021 at 21:35
• then by instantiation we get $| a_{n_2} - a_{n_2 - 1} | < \frac{1}{n_2 -1}$ ( note that since $n_1 \leq n_2 -1 < n_2$ then obviously $n_2 - 1 >0$ ), Note that since $n_2 - 1 \geq 1$ then $| a_{n_2} - a_{n_2 - 1} | < \frac{1}{n_2 -1} \leq 1$, together with the derivation that $| a_{n_2} - a_{n_2 - 1} | > 9$ we get a contradiction. Hence there exists $n_1 \leq n < n_2$ s.t. $a_n \geq 0$. If this isn't wrong, do you have any idea how does this fact help me? Commented Apr 23, 2021 at 21:35

Consider the set $$A$$ of all $$m \in \mathbb{N}$$ such that $$n_1 \leq m < n_2$$ and $$a_{m+1} > 0$$. This set is nonempty since $$m = n_2 - 1 \in A$$. Therefore, it has a minimum element $$n\in A$$ (see the Note below).

Now, if $$a_n > 0$$, then we'd have $$n \neq n_1$$ (since $$a_{n_1} \leq 0$$) so $$n-1 \geq n_1$$ However, then $$n-1 \in A$$ which contradicts the definition of $$n$$ as the minimum element of $$A$$. It follows our supposition that $$a_n > 0$$ must be wrong, so $$a_n \leq 0$$.

To recap, $$n \in A$$ so $$n_1 \leq n < n_2$$ and $$a_{n+1} > 0$$. Also, we just showed that $$a_n \leq 0$$, so this $$n$$ satisfies all the conditions you need.

Note: Here, we used the fact that $$\mathbb{N}$$ is well ordered. In other words, every nonempty subset of $$\mathbb{N}$$ has a minimum element.

This is the fact that justifies "proof by induction", and in fact is equivalent to it. In other words, I could, after some logical manipulation, rewrite the above proof using induction, but I personally prefer the way I've shown. (Both are non-constructive, but the proof by induction somehow feels more non-constructive)

• Somewhat important, but note that the only property of $\mathbb{R}$ we used is that for any $x \in \mathbb{R}$ we either have $x \leq 0$ or $x > 0$. It doesn't even matter that these are mutually exclusive (though they are). That is, the step you were concerned about really does come down to the properties of $\mathbb{N}$, and not the real numbers at all. Commented Apr 23, 2021 at 19:28