A proof from my analysis textbook about convergent sequence in a metric space that is bounded says:
From $$\lim_{n\to \infty} a_n = a$$
$$ (\exists n_0)( \forall n \in N\Bbb)(n \ge n_0 \Rightarrow a_n \in B(a,1)) $$
If $$ n_0 = 1$$ then all terms of a sequence are in open ball B(a,1), so sequence is bounded .
For $$ n_0 \gt 1 $$ then $$ d = \max \{1,d(a,a_1),d(a_1,a_2),\ldots,d(a,a_{n0-1}) \} .$$
Then $$ d(a_n,a_m)\le d(a_n,a) + d(a,a_m)\le 2d$$
so $$ \sup \{d(a_n,a_m) : a_n,a_m \in \{a_n \} \} \le 2d $$
So, a sequence $$\{a_n\}$$ is bounded.
I understood the whole proof. My misunderstanding starts at point where we use triangle inequality to show that a terms out of a ball L(a,1) are in a bigger ball. There I DO NOT understand why we use less OR EQUAL of 2d. In my opinion there should stay ONLY less then 2d, because we talk about open ball, where all terms are inside it.
Where I am wrong?
Thanks in advance!
Best Answer
There is a fundamental flaw in that argument, in that it is not proving what the statement is all about.
Let $(X,d)$ be the metric space. To prove that your sequence is bounded, you must find a $x \in X$ and a $r > 0$ such that for all $n \in \mathbb N$, $d(a_n, x) < r$ (or $\le r$, that is an equivalent statement). So bounding $d(a_n,a_m)$ is essentially pointless.
If you go through the argument until $$ r \overset{def}= \max \{ 1, d(a,a_1),\cdots,d(a,a_{n_0-1}) \} $$ the argument is in fact complete (I used a different letter, $r$, to denote the radius, since $d$ is already used for the metric), since $d(a,a_n) \le r$ for all $n \in \mathbb N$ by definition of $r$! So the sequence is bounded (pick $a$ for the center and $r$ for the radius).
Hope that helps,