Let $S$ be a subset of a metric space $(X,d)$.
I have read (here) the "Sequential characterization of totally bounded subsets" that says the following are equivalent:
1.) $S$ is totally bounded.
2.) every sequence in S has a Cauchy subsequence.
In my class we have proved that the following are equivalent:
$\neg 1$.) $S$ is not totally bounded.
3.) $\exists \epsilon > 0$ and $\{a_n\}_{n \in \mathbb{N}} \subseteq S$ such that $d(a_n, a_m) \geq \epsilon \ \forall n \neq m$
It logically follows that 1.) and 2.) are equivalent to $\neg 3$.), that is the negation of proposition 3.).
But I'm having trouble negating proposition 3.). Here is my attempt
4.) $\forall \epsilon > 0$ and $\forall \{ a_n \}_{n \in \mathbb{N}} \subseteq S$ there exists $n \neq m$ such that $d(a_n, a_m) < \epsilon$.
There has to be something wrong, because that doesn't imply 2.), because for example let $X = \mathbb{R}^2$ and $(b_n) = (n,0)_n = ( (1,0) , (2,0) , (3,0) , (4,0) , \ldots)$ $(c_n)_n = (n, \frac{1}{n})_n = ( (1,1), (2,\frac{1}{2}), (3,\frac{1}{3}), (4,\frac{1}{4}), \ldots)$ and $(a_n) = (b_1, c_1, b_2, c_2, b_3, \ldots) = ((1,0), (1,1), (2,0), (2,\frac{1}{2}), (3,0), (3,\frac{1}{3}), \ldots)$ Then any subsequence of $(a_n)$ is not bounded (and then not Cauchy), but $d(a_n, a_{n+1}) = 1/n$ (with $n$ impair) so it does exists $n \neq m $ such that $d(a_n, a_m) < \epsilon$. What am I doing wrong? Thanks in advance for any help.
Edit: I'm seeing the error here: I have found only one sequence that does not have a cauchy subsequence. What that means is the obvious fact that $\mathbb{R}^2$ is not totally bounded.
So is my negation of 3.) correct?. That would imply, $S$ is totally bounded iff $\forall \epsilon > 0$ and $\forall \{ a_n \}_{n \in \mathbb{N}} \subseteq S$ there exists $n \neq m$ such that $d(a_n, a_m) < \epsilon$
Yes, your negation of $3)$ is correct. You can think of it like this: In a not totally bounded space, there's room for the sequence to "run away". While total boundedness implies the existence of an entire Cauchy subsequence, already the seemingly weaker fact that the sequence has infinitely many pairs of arbitrarily close elements is enough to show that there's no room to "run away".