Could someone check to see if my proof is correct?

Since $a_n→a$, there is an N in the natural numbers such that, if $n≥N $ then$ ,|a_n−a|< \epsilon, $ for $ \epsilon >0$ (Let$\epsilon=a/2$). Additionally, $b_n→b$, there exists an N in the natural numbers such that, if n≥N then $|b_n−b|< \epsilon$, for $\epsilon >0$ (Let$\epsilon=b/2$).Thus , $a−a/2 < a_n< a+a/2$ and $b−b/2< b_n < b+b/2$. Since $a_n≤b_n$, it follows that $a−a/2≤b−b/2$. Thus,$a≤b$. QED


1 Answer 1


There are a few minor problems that are distracting, and it's a good habit to fix things like that first so that you can find other errors, if there are any other errors.

First, after you assign a value to epsilon in terms of "a", you get a value N, but re-use the same variable name "N" when you are working with "b", although you don't know that it has the same value as the original N.

Second, you re-use the name $\epsilon$ when working with "b", just as you re-used the name N.

Third, it's not clear how you obtain $b−b/2< b_n$ unless you are assuming that b is greater than or equal to zero. However, there could be an easy fix: replace b/2 with the absolute value of b/2. Usually, it's better if you look for a remedy for your first draft proof after somebody else points out a potential gap or flaw in your reasoning.

Now, your inequalities involving $a_n $ and $ b_n $ are being combined, so you need n to be the same value in those inequalities, and that means that you want to select n big enough to be bigger than both of the N values. So you should explicitly say something about the values of n for which your inequalities are asserted to be true.

If I can make any of that clearer, please let me know. Given my role as nitpicker in this answer, I accept an obligation to be very clear.

Despite all of the complaints above, what you have done is pretty good, especially because the problem looks more demanding than interesting. I would save your first draft, as a reminder that it's easier to get everything perfect if you allow some imperfection in the first draft. It's likely more interesting to you than something in a textbook, because you wrote it yourself.

Wait, I see another potential problem: you don't say specifically what inequalities you combine or how you combine them to justify asserting "Since $ a_n≤b_n$ it follows that." Are you using something of the form (x <= y and y <= z) therefore (x <= z)? You have to show your work there step-by-step, because otherwise you could be putting the burden on the reader to fill in the gap that shows that it follows. You know what you are supposed to obtain, but you have actually get there one step at a time. So, maybe it's now a more interesting problem than it initially seemed to be!

  • $\begingroup$ Thank you for your critiques! I appreciate any and all constructive criticism. $\endgroup$ Oct 24, 2019 at 2:27

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