I'm trying to understand the proof of Brouwer's Fixed Point Theorem using Sperner's Lemma. For example, pg.10-11 in A Combinatorial Approach to The Brouwer Fixed Point Theorem.
I was able to follow it until it took the limit of the convergent sequence. Why would the limit give "less than or equal to" instead of strict inequality? Most of such proofs say this is due to continuity. Because my lack of knowledge on analysis, this is not clear to me. Did this occur as a special case or a general result? Please explain. Thanks!
Limits preserve weak inequalities, not strict inequalities. The reason for this is that $(0, \infty)$ is not closed in $\Bbb{R}$, but $[0, \infty)$ is. That is, if $x_n \in [0, \infty)$ that converges to $x$, then $x \in [0, \infty)$. If we simply take $x_n \in (0, \infty)$, we can conclude $x \in [0, \infty)$, but it is possible for $x = 0$, e.g. if $x_n = \frac{1}{n}$.
So, if $x_n > y_n$ while $x_n \to x$ and $y_n \to y$, we see that $x_n - y_n \in (0, \infty)$ converges to $x - y$. We do get $x - y \in [0, \infty)$, i.e. $x \ge y$, but we may not have $x - y \in (0, \infty)$, i.e. $x > y$. That is, inequality is preserved, but not strict inequality.