In a proof of the completeness of $l^\infty$ on this page, the author finds a candidate limit point $x$ for a cauchy seqeuence $x^n$. He then chooses an $\varepsilon$ and then considers $\varepsilon/2$ and says we can find $N > 0$ such that $$|x_k^n - x_k^m | < \frac{\varepsilon}{2},$$ for all $k$ and for all $n, m > N$.
Then he takes the limit as $m \to \infty$ and obtains $$|x_k^n - x_k| \le \frac{\varepsilon}{2},$$ for all k and for $n > N$.
Question
I don't see how this has become a less than or equal to expression as opposed to the original less than expression? Here are my detailed workings:
For $n, m > N$ we have
$ \begin{align} |x_k^n - x_k^m| & < \frac{\varepsilon}{2} \\ \lim_{m \to \infty} |x_k^n - x_k^m| & < \lim_{m \to \infty} \frac{\varepsilon}{2} \\ |x_k^n - \lim_{m \to \infty} x_k^m| & < \frac{\varepsilon}{2} \\ |x_k^n - x_k| & < \frac{\varepsilon}{2} \\ \end{align} $
Am I missing something or is the author incorrect? Where in my workings have I make a mistake if I am indeed incorrect?
The author is correct. Sadly, strict inequalities are not conserved when taking the limit. They become weak inequalities in general.
Consider a simple example: the sequence $(a_n)_{n\geq 1}$ defined by $a_n = \frac{1}{n}$. Clearly, we have $a_n > 0$ for all $n\geq 1$. Yet, taking the limit, $$ \lim_{n\to\infty} a_n = 0 $$ so the strict inequality "dies."
You can see it happen in related occurrences, maybe simpler to understand — for instance with the supremum of a set. The supremum of $S=[0,1)$ is $1$, even though every element $s\in S$ is strictly smaller than $1$. This is exactly the same phenomenon: if your sequence has elements getting arbitrarily "close to making the strict inequality tight," then the limit cannot satisfy the strict inequality as well. Otherwise, there would still be elements of the sequence staying bounded away from the limit ("close to" the RHS of the inequality).