Question 3.T.5: Suppose xn → x and yn → y. If xn ≤ yn for all n, then x ≤ y.

Let ε be any positive number. Since $x_{n} → x$ and $y_{n} → y$, there exist $N_{1}, N_{2} ∈ \mathbb{N}$ such that

$|x_{n} − x| < \frac{ε}{2}$  for all $n ≥ N_{1}$

$|y_{n} − y| < \frac{ε}{2}$  for all $n ≥ N_{2}$.

Thus, for all n ≥ max$\left\{N_{1}, N_{2}\right\},$ we have

$x − \frac{ε}{2} < x_{n} < x + \frac{ε}{2}$

$y − \frac{ε}{2} < y_{n} < y + \frac{ε}{2}$.

Using $x_{n} ≤ y_{n}$, we arrive at

$x − \frac{ε}{2} < x_{n} ≤ y_{n} < y + \frac{ε}{2},$

which implies

x < y + ε.

But since ε > 0 is arbitrary, we must have x ≤ y (Exercise 2.2.2).