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).