Show that if $a_{n} \geq 0$ for all $n \in \mathbb{N}$ then $a \geq 0$ - Proof Explanation

Question

Assume the sequence $a_{n} \rightarrow a$. Show that if $a_{n} \geq 0$ for all $n \in \mathbb{N}$ then $a \geq 0$

I've gone over some of the other posts about it, and I have worked out the solution "somewhat". Where I am having problems is actually obtaining the contradiction myself formally. So as a template of the proof:

Proof:

Suppose towards contradiction $a < 0$

Consider $\epsilon = |a|$. By the definition of convergence there exists a $N > 0$ such that for all $n \geq N$, $|a_{n} - a| < \epsilon = |a|$.

Expanding this I will arrive at: $$-|a| < a_{n} - a < |a|$$

It is here where I get stuck. I know that when when it is worked out I will get a situation where $a_{n} < 0$ and that will be the contradiction, but arriving at that step by doing the steps is getting more difficult than it really should be......assitance?

Answer 1

Think of it geometrically. If $a< 0$ then $a$ is a distance of $|a|$ to the "left" of $0$. And as all $a_n > 0$ all $a_n$ are to the "right" of $0$. And the distance between them, $|a_n - a|$, is larger than the distance from $a$ to $0$ which is a positive $|a|$. $|a_n - a|$ can never be less than $|a|$.

So how to do it analytically where we can't rely on geometric "common sense"?

Well since $a < 0 \le a_n$ we have $a_n > a$ so $a_n - a > 0$ so $|a_n - a| = a_n - a$. And as $a < 0$ $a_n - a = a_n + |a|$. And as $a_n \ge 0$ and $a_n + |a| \ge |a| > 0$.

So IF $a< 0$ and $a_n \ge 0$ then $|a_n -a| \ge |a| > 0$.

So for any $0< \epsilon \le |a|$ it can not be true for any $a_n > 0$ that $|a_n - a| < \epsilon$.

Answer 2

You supposed $a<0$, hence $|a|=-a$. So from $-|a|<a_n-a<|a|$ we get $a_n<|a|+a=-a+a=0$ which is a contradiction.

Answer 3

Question: Assume the sequence $a_n→a$. Show that if $a_n≥0$ for all $n∈\mathbb{N}$ then $a≥0$.

Answer: If possible let the conclusion is not true and $a\lt 0$.

Then since $\{a_n\}$ converges to $a$, so for $\quad\epsilon =-a \gt 0,\quad \exists\quad k\in\mathbb{N}\quad$such that $$|a_n-a|\lt\epsilon\qquad \forall\quad n\ge k$$ $$\implies a-\epsilon\lt a_n\lt a+\epsilon\qquad \forall\quad n\ge k$$ $$2a\lt a_n \lt 0 \qquad \forall\quad n\ge k$$

But given that $\quad a_n\ge 0 \quad$, so our assumption is wrong.

Hence$\quad a=\lim a_n \ge 0 \quad$

Answer 4

This is tantamount to showing that $[0,\infty)$ is closed, or equivalently, that $(-\infty,0)$ is open.

It is clear that if $a \in (-\infty,0)$ (or $a<0$) that there is some $\epsilon>0$ such that $B(a,\epsilon) \subset (-\infty,0)$.

Hence if $a_n \to a$, then after some $N$ we see that $a_n \le a+{1 \over 2}\epsilon < 0$.

Hence if $a_n \ge 0$ and $a_n \to a$ we must have $a \ge 0$.

Answer 5

Assume $a <0$.

Choose $\epsilon=|a|/2$.

Consider

$B_{\epsilon}(a)=$

{$x|a-\epsilon <x<a+\epsilon= a/2$}.

$B_{\epsilon}(a)\cap${$a_n| n \in \mathbb{N}$}$=\emptyset$. ($a_n \ge 0$)

Contradiction to $\lim a_n \rightarrow 0$.