Exercise 5.2.2

March 25, 2020 Issa Rice 13 Comments

Exercise statement

Let $\varepsilon > 0$ . Show that if $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are eventually $\varepsilon$ -close, then $(a_n)_{n=1}^\infty$ is bounded if and only if $(b_n)_{n=1}^\infty$ is bounded.

Hints

Use the triangle inequality.

How to think about the exercise

This exercise is similar to both Exercise 5.1.1 and Exercise 5.2.1. I don’t have anything to add beyond what I said in those posts.

Model solution

Suppose $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are eventually $\varepsilon$ -close, and suppose $(a_n)_{n=1}^\infty$ is bounded. Since $(a_n)_{n=1}^\infty$ is bounded, there exists a number $M_1 \geq 0$ such that $|a_n| \leq M_1$ for all $n \geq 1$ . Since $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are eventually $\varepsilon$ -close, there exists $N \geq 1$ such that $|a_n - b_n| \leq \varepsilon$ for all $n \geq N$ . By the triangle inequality we have $|b_n| - |a_n| \leq |a_n - b_n|$ (to get this, you can start with the ordinary triangle inequality $|x+y| \leq |x| + |y|$ and then substitute $x:= a_n$ and $y := b_n - a_n$ ), so $|b_n| \leq |a_n| + \varepsilon \leq M_1 + \varepsilon$ for all $n \geq N$ . This bounds the sequence $(b_n)_{n=N}^\infty$ (i.e. the infinite tail of $(b_n)_{n=1}^\infty$ ). By Lemma 5.1.14, the finite sequence $b_1, \ldots, b_{N-1}$ is also bounded, say by the number $M_2 \geq 0$ . Thus we can take $M := M_1 + M_2 + \varepsilon$ as a bound on $(b_n)_{n=1}^\infty$ .

For the other direction, just interchange $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ in the above.

13 thoughts on “Exercise 5.2.2”

Diego says:

May 23, 2022 at 3:43 pm

Do you think it would be possible to pick an arbitrary $\epsilon$ ? I proved it letting $\epsilon =1$ . The reasoning is all the same, but you end up with $|b_n| \leq M_1 + M_2 + 1$ .

I don’t think it should matter as you are saying that it is still bounded, but just picking bigger $M$ s ; i.e., $M$ bigger than the $N$ such that the distance between $a_n$ and $b_n$ is 1.

Also, how did you write $|b_n| - |a_n| \leq |a_n-b_n|$ ? By strictly applying $|x+y| \leq |x| + |y|$ you shouldn’t get any minus signs right? I mean I still got it to work following an approach similar to 5.2.1 and writing $|b_n| = |b_n + a_n - a_n|$ and then, applying the triangle inequality.

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1. Issa Rice says:
  
  May 23, 2022 at 7:57 pm
  
  You are not allowed to take $\varepsilon$ to be a specific number like $1$ , since the exercise is to show that the claim works for an arbitrary $\varepsilon$ (that’s what the “Let $\varepsilon > 0$ ” bit is doing in the exercise statement).
  
  To get $|b_n| - |a_n| \leq |a_n - b_n|$ , you can start with the ordinary triangle inequality $|x+y| \leq |x| + |y|$ and then substitute $x:= a_n$ and $y := b_n - a_n$ . I’ll edit the post to add this reasoning in; thanks for bringing it to my attention.
  
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Diego says:

May 24, 2022 at 11:53 am

Ok, the triangle inequality is clearly fine the way you did it. However, I am still confuse from the first bit.

Indeed one thing that confused me was why the exercise stated: “Let $\epsilon > 0$ . Show that if $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are eventually $\epsilon$ -steady … Why write this instead of writing that the two sequences are equivalent? After all if they are equivalent they are clearly eventually $\epsilon$ -steady for all $\epsilon > 0$ . What is the purpose of writing the sentence as Tao did? It seems to make sense only if the two sequences are not equivalent; i.e., if there is an $\epsilon >$ such that the two sequences do not get closer than that. Maybe I am missing something here?

On a separate note, I am still not happy as to why letting $\epsilon = 1$ does not work. After all if choose an $N_1$ such that the distance between $a_n$ and $b_n$ s is less than 1, and note that all $a_n$ s are bounded by some $M$ , then all the $b_n$ s must also be bounded by the same $M$ plus the maximum distance they have from the $a_n$ s which is $1$ for that particular $N$ . I don’t see what is the problem with this reasoning. At the end I want to prove that the $b$ sequence is bounded, and I do conclude that. Obviously this is nitty gritty details, as I understand the proof with an arbitrary $\epsilon > 0$ , but it just seems that it should also work for a given $\epsilon$ . Let me know what you think.

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1. Issa Rice says:
  
  May 24, 2022 at 11:05 pm
  
  I thought about this some more and have a better idea of what the confusion might be about.
  
  If we rewrite the exercise statement using first-order logic notation, we get:
  
  $\forall \varepsilon (P(\varepsilon) \implies Q)$
  
  where the domain of quantification is the set $\mathbf Q^+$ of positive rational numbers, $P(\varepsilon)$ is the statement that the two sequences are eventually $\varepsilon$ -close, and $Q$ is the statement that the two sequences are either both bounded or both unbounded.
  
  This is not the same as $(\forall \varepsilon P(\varepsilon)) \implies Q$ , which is the statement “if the two sequences are equivalent, then one is bounded iff the other one is”.
  
  The above statement can be rewritten in a different way:
  
  $\forall \varepsilon (P(\varepsilon) \implies Q)$
  is equivalent to
  $\forall \varepsilon (\lnot P(\varepsilon) \lor Q)$
  is equivalent to
  $(\forall \varepsilon \lnot P(\varepsilon)) \lor Q$
  is equivalent to
  $(\lnot \exists \varepsilon P(\varepsilon)) \lor Q$
  is equivalent to
  $(\exists \varepsilon P(\varepsilon)) \implies Q$
  
  (Let me know if any of the above steps don’t make sense, and I can explain.)
  
  The last statement above is still not the statement you showed: if one wants to prove $(\exists \varepsilon P(\varepsilon)) \implies Q$ and assumes the hypothesis $\exists \varepsilon P(\varepsilon)$ , one must work with the assumption that there is some non-specific $\varepsilon$ that makes $P(\varepsilon)$ true, rather than setting $\varepsilon$ to be some concrete value.
  
  The statement you proved is $P(1) \implies Q$ . This establishes $\exists \varepsilon (P(\varepsilon) \implies Q)$ . By doing logic manipulations similar to the above we can show that this is logically equivalent to $(\forall \varepsilon P(\varepsilon)) \implies Q$ . So actually, you ended up showing “if the two sequences are equivalent, then one is bounded iff the other one is”.
  
  As long as the domain of quantification, namely $\mathbf Q^+$ , is non-empty, $\forall \varepsilon (P(\varepsilon) \implies Q)$ (or equivalently $(\exists \varepsilon P(\varepsilon)) \implies Q$ ) implies $\exists \varepsilon (P(\varepsilon) \implies Q)$ (or equivalently $(\forall \varepsilon P(\varepsilon)) \implies Q$ ). So the exercise statement implies what you proved, but the reverse is not true.
  
  You are correct though that if all we care about is proving $Q$ , then showing $P(1)$ is sufficient.
  
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2. Issa Rice says:
  
  May 24, 2022 at 11:40 pm
  
  Here’s another way to think about this. In general, when proving a statement of the form $P \implies Q$ , the weaker the hypothesis statement $P$ is, the stronger the overall implication is. So in this problem, $\forall \varepsilon P(\varepsilon)$ is the strongest hypothesis, because it implies $P(1)$ and $\exists \varepsilon P(\varepsilon)$ . But if we compare $P(1)$ and $\exists \varepsilon P(\varepsilon)$ , then actually $P(1)$ is stronger because it implies $\exists \varepsilon P(\varepsilon)$ (but not the other way around). So this exercise is getting us to prove $Q$ starting from the weakest of these three hypotheses, which results in the strongest implication, which is the most “interesting” statement.
  
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Diego says:

May 26, 2022 at 6:47 pm

Well, I must say that I agree with you now. I think that what I did is misread Tao’s question. In particular I was thinking of

$[\forall \epsilon P(\epsilon)] \implies Q$

However, Tao most probably does not mean this as this would mean (as you correctly point out) that the two sequences are equivalent. However, he is very careful in not stating that.

Now it also much clearer why $P(1)$ is clearly sufficient to show $Q$ , it is simply not sufficient to show $\forall \epsilon (P(\epsilon) \implies Q)$ , which makes sense.

This was a rather insightful discussion. Thanks a lot.

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1. Issa Rice says:
  
  May 26, 2022 at 9:17 pm
  
  Glad I was able to help! 🙂
  
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deepaksm says:

April 6, 2023 at 2:21 pm

Thanks for the solution.

Can you please say why we’re showing the below in the proof?
“By Lemma 5.1.14, the finite sequence $b_1, \ldots, b_{N-1}$ is also bounded, say by the number $M_2 \geq 0$ . Thus we can take $M := M_1 + M_2 + \varepsilon$ as a bound on $(b_n)_{n=1}^\infty$ .”

especially why do we need two M’s to show that it’s bounded since we can easily say that it is bounded because it’s finite?

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1. Issa Rice says:
  
  April 6, 2023 at 9:38 pm
  
  One part of the sequence is bounded because it’s finite. Another part of the sequence is bounded because we proved some bound for it. How do we know that the entire sequence is bounded? The definition of “bounded sequence” does not allow you to say “well, one part is bounded by one thing, and another part is bounded by another thing, so the whole sequence is bounded”. Instead, you must produce a single bound for the entire sequence. There are multiple ways of producing such a bound, but some common ways are to add the two separate bounds or to take the max of the separate bounds.
  
  I’m not entirely sure I answered your question. If I didn’t, please say some more words about what seems confusing.
  
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2. deepaksm says:
  
  April 7, 2023 at 5:14 am
  
  After some thought, now I guess I understand.
  Though we’d found that $|b_n| \leq |a_n| + \varepsilon \leq M_1 + \varepsilon$ , there’s no guarantee that $n$ starts at zero, which imply the existence of a finite sequence of $b_k$ where $0 \leq k < n$ . if at all such finite sequence exists, then by lemma 5.1.14, it is bounded (by some $M_2$ ).
  From here, like you said, we can take either $\max(M_1, M_2) + \varepsilon$ (or) $M_1 + M_2 + \varepsilon$ as the bound for the entire $b_n$ sequence.
  
  Please let me know if this reasoning is correct.
  
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3. Issa Rice says:
  
  April 7, 2023 at 7:20 am
  
  That sounds right, except that $|b_n| \leq |a_n| + \varepsilon \leq M_1 + \varepsilon$ holds for all $n \geq N$ . So it is $N$ (rather than $n$ ) that we can’t be sure is $1$ (the sequences in this exercise start at $1$ rather than zero).
  
  For the max bound, using $\max(M_1 + \varepsilon, M_2)$ would be semantically clearer (we are saying “make it a number that’s as big as each of the two bounds we found”), although the bound you gave also works.
  
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4. deepaksm says:
  
  April 7, 2023 at 7:30 am
  
  Got it, I actually found those errors (n starts at 1 and not at 0) just after submitting the reply but couldn’t edit it. Also, I see that the latex is not rendering properly for some time, I wonder whether that’s usual or I’d need to add the latex quote – “ around the symbols.
  
  However, your answer clarifies everything.
  Thanks for your time! 🙂
  
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5. Issa Rice says:
  
  April 8, 2023 at 6:53 pm
  
  That happens because I manually fix up the LaTeX, since otherwise it’s hard for me to read the comment 😛
  
  On WordPress blogs like this one, you can write LaTeX by writing $latex \sin x$ for example (i.e. like normal LaTeX except that you have to also write “latex” after the beginning dollar sign). You can find more documentation here.
  
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