# Proof

By assumption, $$(a_n)_{n\in\mathbb N}$$ and $$(b_n)_{n\in\mathbb N}$$ are convergent real sequences with the limits $$\lim_{n\rightarrow\infty} a_n=a$$ and $$\lim_{n\rightarrow\infty} b_n=b$$. We have to show the following:

### $$( 1 )$$ If $$a_n \le b_n$$ for all $$n\in\mathbb N$$, then $$a \le b$$.

Using the definition of the order relation for real numbers, the relations "$$\le$$" and "$$>$$" are mutually exclusive. Thus, let assume that $$a_n \le b_n$$ for all $$n\in\mathbb N$$, but $$a > b$$. Set $$\epsilon:=(a-b)/2 > 0$$. Since there are indices $$N_a(\epsilon),N_b(\epsilon)\in\mathbb N$$ such that

$|a_n-a| < \epsilon~\forall n\ge N_a(\epsilon)\quad\quad\text{and}\quad\quad|b_n-b| < \epsilon~\forall n\ge N_b(\epsilon),$

then, for all $$n\ge\max(N_a(\epsilon),N_b(\epsilon))$$, it follows

$a_n > a - \epsilon\quad\quad\text{and}\quad\quad b_n < b + \epsilon .$

By construction, we have $$a - \epsilon=b +\epsilon$$, so it follows $$a_n > b_n$$ for all $$n\ge \max(N_a(\epsilon),N_b(\epsilon))$$. But $$a_n \le b_n$$ by assumption, which is a contradiction.

### $$( 2 )$$ From $$a_n < b_n$$ for all $$n\in\mathbb N$$ it does not (!) follow that $$a < b$$.

It is sufficient to find only one counterexample, for which $$a_n < b_n$$ for all $$n\in\mathbb N$$, but $$a \not < b$$. This counterexample can be given for the sequences $$a_n = 0$$ and $$b_n=1/n$$ for all $$n\in\mathbb N$$, for which we have

$\lim_{n\rightarrow\infty} a_n=\lim_{n\rightarrow\infty} 0= 0 = \lim_{n\rightarrow\infty} \frac 1n=\lim_{n\rightarrow\infty} b_n$

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983