If a_n < b_n for all n, then lim a_n < lim b_n

[demonelite123]

If $a_n$ and $b_n$ are convergent sequences and $a_n ≤ b_n$ for all n, show that a ≤ b where a and b are the limits of $a_n$ and $b_n$ respectively.

since a_n and b_n are convergent, there exists an N1 such that $|a_n - a| < ε$ for all n > N1 and an N2 such that $|bn - b| < ε$ for all n > N2. I then choose N = max(N1, N2) so for all n > N, the 2 inequalities are satisfied. Since i want to show that a ≤ b, i take $a < a_n + ε ≤ b_n + ε$, but i am stuck here since b_n + ε is not less than b. Since this leads to a dead end, can someone give me a hint on how to approach this problem?

 

[voko]

Prove by contradiction that a > b is impossible.

 

[jbunniii]

Hint: if $a > b$, then you can find an $\epsilon > 0$ such that $a - \epsilon > b + \epsilon$.

 

[Dick]

If $a_n$ and $b_n$ are convergent sequences and $a_n ≤ b_n$ for all n, show that a ≤ b where a and b are the limits of $a_n$ and $b_n$ respectively.

since a_n and b_n are convergent, there exists an N1 such that $|a_n - a| < ε$ for all n > N1 and an N2 such that $|bn - b| < ε$ for all n > N2. I then choose N = max(N1, N2) so for all n > N, the 2 inequalities are satisfied. Since i want to show that a ≤ b, i take $a < a_n + ε ≤ b_n + ε$, but i am stuck here since b_n + ε is not less than b. Since this leads to a dead end, can someone give me a hint on how to approach this problem?

Do a proof by contradiction. Assume that a>b. Pick ε=(a-b)/4. Doesn't that lead to a contradiction?

 

[demonelite123]

thanks for your replies. i am working on it but still having trouble finding a contradiction. i know that a > b and a_n ≤ b_n along with $|a_n - a| < ε$ and $|b_n - b| ≤ ε$ will lead me to a contradiction, but i have been manipulating these all together and still can't find the contradiction.

i've also tried setting ε = (a-b)/4 like Dick said, but I'm not sure of how to get a contradiction from that either. i get that a_n < (5a-b)/4 and b_n < (a + 3b)/4 which doesn't tell me much. can you explain a little more about how the contradiction arises in this case?

 

[voko]

Assuming $a > b$, and choosing $\epsilon$ so that $a - \epsilon > b + \epsilon > b_n$, $a_n > a - \epsilon > b + \epsilon > b_n$, which contradicts $a_n \le b_n$.

 

[demonelite123]

thanks for your reply voko!

i know that you told me to do that earlier but i wasn't sure if that in itself had to be proved or not. Is that step considered elementary or does it actually need to be rigorously proved as well?

i was trying to do this problem using only the inequalities given and using only operations/properties proved true for inequalities such as adding/subtracting things on both sides. but like i said earlier, i wasn't able to get very far.

 

[voko]

i know that you told me to do that earlier but i wasn't sure if that in itself had to be proved or not. Is that step considered elementary or does it actually need to be rigorously proved as well?

Since you are requested to prove that, it is best not to make any assumptions as to the rigor. The sketch I gave above is still not completely formal, since one needs to prove that it is possible to choose $\epsilon$ with the required property (even though it might seem obvious). Dick suggested one particular way, but it is still up to you to prove that such an $\epsilon$ satisfies $a - \epsilon > b + \epsilon$.