How should I complete this argument?

Question

I am trying to prove the following proposition.

Proposition. Given that $(a_n)\to a$. If every $(a_n)$ is an upper bound for a set $B$, then $a$ is also an upper bound for $B$.

Proof. We prove the contrapositive. Assume that $a<b$ for some $b\in B$, then by hypothesis there exists an $N\in\mathbf{N}$ such that $|a_n-a|<\frac{|b-a|}{2},\forall n\ge N$.

Now geometrically I can see that this should imply that $a_N<b$ but I can't seem to prove this any suggestions?

Answer 1

Assume that $a<b$ for some $b\in B$. Call $\delta=b-a>0$. Then, by convergence there exists $n_0\in \mathbb{N}$ such that $|a-a_n|<\delta$ for $n\geq n_0$. If $a_n\leq a$ for some $n$ then you're done because $a_n\leq a< b$. If not, then $a_n-a<b-a$, and then $a_n<b$ for every $n$ big enough.

Answer 2

Let $b \in B$, since every $a_n$ is an upper bound of $B$, we get $b \le a_n$ for all $n$. This gives (with $n \to \infty$):

$b \le a$. Thus we have proved:

for every $b \in B$ we have $b \le a$.

Answer 3

Consider the sequences $a_n$, $b_n=b$.

$x_n:= a_n-b_n \ge 0$, $n\in \mathbb{Z^+}$.

$x_n \ge 0$ converges to $x:= a-b$.

Need to show that $x \ge 0$.

Assume $x <0$.

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

There is a $n_0$ such that for $n\ge n_0$

$|x_n-x| \lt \epsilon =|x|/2$, or

$-|x|/2 +x \lt x_n \lt x +|x|/2 \lt 0$.

Contradiction to $x_n \ge 0$.