How to prove this property of limit of sequences

Question

I read a property about limit of sequences as:

suppose $ a_{n} \leq b_{n}, \lim _{n \rightarrow \infty } a_{n}=a, \lim _{n \rightarrow \infty } b_{n}=b $, then $a \leq b$.

I know a proof by contradiction and I am wondering how to prove this property directly? Thanks a lot!

score 3 · Accepted Answer · answered Nov 29 '13 at 14:40

Choose $\epsilon > 0$, then there is $N_0 \in \mathbb{N}$ such that $$ b_n < b+\epsilon \quad\forall n\geq N_0 $$ and $N_1 \in \mathbb{N}$ such that $$ a-\epsilon < a_n \quad\forall n\geq N_1 $$ So for $n= \max\{N_0, N_1\}$, one has $$ a-\epsilon < a_n \leq b_n < b+\epsilon $$ and so $$ a < b+2\epsilon \quad\forall \epsilon > 0 $$ and so $a\leq b$.

However, this last statement is usually proved using contradiction, so I'm not really sure if this satisfies your (rather stringent) conditions.

thanks a lot! I really appreciate your help and it does inspire me a lot! thank you! — Brain Zhang, Nov 30 '13 at 02:43

score 2 · Answer 2 · answered Nov 29 '13 at 14:46

2

$$b-a=\lim_{n\to\infty}b_n-\lim_{n\to\infty}a_n=\lim_{n\to\infty}(b_n-a_n)\geq\lim_{n\to\infty}0=0$$

answered Nov 29 '13 at 14:46

Shuchang

9,800

This has been flagged as "low quality". Why? This is a great answer. – Ahaan S. Rungta Nov 29 '13 at 15:05
1

@AhaanRungta Auto-flagged by the system, no text, just a formula, that makes it suspicious. But sometimes answers which meet these criteria are okay or even good, that's why the system puts it in the queue, so that people look at it. – Daniel Fischer Nov 29 '13 at 18:48
@DanielFischer Ah, I didn't know that. Thanks! – Ahaan S. Rungta Nov 29 '13 at 18:49
@AhaanRungta Hmm.. Why not help upvote since it's not disappointing. – Shuchang Nov 30 '13 at 00:12
@ShuchangZhang My voting was locked at the time. Done now. – Ahaan S. Rungta Nov 30 '13 at 00:24
@AhaanRungta Oh, thanks. Even I didn't know my post was locked. The system should have at least a signal for that. – Shuchang Nov 30 '13 at 00:27
@ShuchangZhang No, your post wasn't locked, my voting for the day was locked because I used the daily maximum of 40 votes. – Ahaan S. Rungta Nov 30 '13 at 00:47
@AhaanRungta Well, but I still didn't know it was flagged. – Shuchang Nov 30 '13 at 00:51
@ShuchangZhang See DanielFischer's response above. – Ahaan S. Rungta Nov 30 '13 at 00:52
@AhaanRungta I did but I think there shall be a reminder whenever people (or machine) do something to you. – Shuchang Nov 30 '13 at 01:01
@ShuchangZhang Wait, what? Sorry, I don't understand. – Ahaan S. Rungta Nov 30 '13 at 01:02
@AhaanRungta I mean the system should tell me immediately after flagging me. But obviously not. – Shuchang Nov 30 '13 at 01:12
@ShuchangZhang I have a question on the last inequality. Do we get this inequality from the fact that $ 0\leq b_{n}-a_{n}$ and then $\lim 0 \leq \lim (b_{n}-a_{n})$? but is this property just what we want to prove? – Brain Zhang Nov 30 '13 at 02:45
@MartinTien Don't think like that. Apply monotonic property, $0\leq b_n-a_n$ implies $0\leq\liminf(b_n-a_n)\leq\lim(b_n-a_n)$ – Shuchang Nov 30 '13 at 06:12

How to prove this property of limit of sequences

2 Answers2