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Problem: Suppose $a_n \to a$ and $a_n \geq b$ for all $n$. Show that $a \geq b$.

My proof: Assume that $b > a$ and let $\epsilon = b - a$. By definition if $\{a_n\}$ converges then $|a_n - a| < \epsilon$ for $n \geq N$. By expanding the absolute value, I see that $a_n - a < b- a$. By adding $a$ to both sides, we get that $a_n < b$ a contradiction.

Is my proof correct? And is there any way to tackle a problem like this without constructing an $\epsilon$?

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    Do you have the Monotonicity Theorem to work with? It says $a_n\ge b_n$ for all $n$ implies $\lim a_n \ge \lim b_n$. That would make quick work of it.2012-03-31

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Your idea is correct, but you’ve not handled the absolute value carefully enough, and the wording could be considerably improved. For instance, you should say where $N$ comes from. Here’s a better version that sticks reasonably closely to what you wrote.

Assume that $b>a$, and let $\epsilon=b-a>0$. Since $a_n\to a$, there must be some positive integer $N$ such that $|a_n-a|<\epsilon$ whenever $n\ge N$. In particular, $|a_N-a|. There are now two possibilities. If $a_N\ge a$, then $|a_N-a|=a_N-a$, and we can add $a$ to both sides to get $a_N, a contradiction. And if $a_N, then $a_N, with the same contradiction. Thus, it’s impossible that $b>a$, and hence $a\ge b$.