(related to Proposition: How Convergence Preserves the Order Relation of Sequence Members)
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:
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.
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
\[\lim_{n\rightarrow\infty} a_n=\lim_{n\rightarrow\infty} 0= 0 = \lim_{n\rightarrow\infty} \frac 1n=\lim_{n\rightarrow\infty} b_n\]
