Prove that if $x_n \leq y_n, \forall n$ then the same inequality holds for their limits, i.e. $x\leq y$

Question

Suppose $\{x_n\}\: \text{and}\: \{y_n\}$ are two sequences s.t. $$\lim_{n\to\infty}x_n=x$$ and $$\lim_{n\to\infty}y_n=y.$$ Then if $x_n \leq y_n, \forall n$, then $x \leq y$.

Proof: Fix $\epsilon >0.$ Observe that $\exists N$ s.t. if $n\geq N$ then $x_n\leq y_n$ and $$|x-x_n|+|y-y_n|<\epsilon/2. \quad{\text{(1)}}$$ Then if $n>N$, it follows that, $$x-y\leq x_n+\epsilon/2-(y_n-\epsilon/2)=x_n-y_n+\epsilon\leq \epsilon. \quad{\text{(*)}}$$ Since $\epsilon$ was arbitrary we have, $x-y\leq 0$ $\implies x \leq y.$

I don't get how step (*) was set up using (1). Any thoughts?

Answer 1

Basically they use (1) in the follwoing two forms:

• $-\frac{\epsilon}{2} < \color{green}{x-x_n < \frac{\epsilon}{2}} \Rightarrow \color{green}{x < x_n + \frac{\epsilon}{2}} $
• $-\frac{\epsilon}{2} < y-y_n < \frac{\epsilon}{2} \Leftrightarrow\color{green}{\frac{\epsilon}{2} > -y+y_n} > -\frac{\epsilon}{2}\Rightarrow \color{green}{-y < \frac{\epsilon}{2} - y_n} = \color{green}{-(y_n - \frac{\epsilon}{2}) }$

Now, you get adding the green parts: $$x-y < x_n + \frac{\epsilon}{2} - (y_n - \frac{\epsilon}{2})$$

Answer 2

$$|x-x_n | + |y-y_n| < \frac{\epsilon}2,$$

then we have $$|x-x_n | < \frac{\epsilon}2, |y-y_n | < \frac{\epsilon}2$$

which implies that $$x < x_n + \frac{\epsilon}2, y >y_n - \frac{\epsilon}2$$

$$x < x_n + \frac{\epsilon}2, -y < -y_n + \frac{\epsilon}2$$

Now, sum up the two inequalities,

$$x -y < x_n + \frac{\epsilon}2-y_n + \frac{\epsilon}2=x_n - y_n + \epsilon$$

Answer 3

An option .

$z_n= y_n-x_n \ge 0$,. $z:= y-x$.

Show that $z \ge 0$.

Assume $z \lt 0$.

Choose $\epsilon = -(z/2) \gt 0$.

There is a $n _0 \in \mathbb{Z^+}$ such that for $n\ge n_0$

$ |z_n-z| \lt \epsilon.$

$-\epsilon +z \lt z_n \lt \epsilon +z = z/2 \lt 0$.

Contradiction, since $z_n \ge 0$.