I am trying to prove that a compact subset, $K$, of a metric space $(X,d)$ is bounded, using the sequential definition of compactness. So far I have the following, but am unsure about where to go next. Am I at least on the right track?
I decided to proceed by contradiction, assuming that $K$ is unbounded (and arriving at a contradiction of the fact that $K$ is compact). I first tried to construct a sequence in $K$, using the fact of $K$'s unboundedness, which I conjectured would not have a convergent subsequence. I constructed it as follows.
If $K$ is unbounded, then, for any $M>0$, there exists $x,y\in K$ s.t. $d(x,y)>M$. As such, we may construct the following sequence, in $K$. Let $a_{1}, a_{2}$ be chosen s.t. $d(a_{1}, a_{2})>1$. Generally, for each $n \in \{1,3,5,...\}$, let $a_{n}, a_{n+1}$ be chosen such that $d(a_{n}, a_{n+1})>n$. This completes the construction of $\{a_{n}\}$, which we now prove does not satisfy the Cauchy Criterion.
Consider some arbitrary $\epsilon>0$, and $N \in\mathbb{N}$. Choose $j\in \{1, 3, 5,...\}$ such that $j>\max\{N, \epsilon\}$. Then, set $n=j, m=j+1$. We have that $d(a_{n}, a_{m})=n>\epsilon$, where $m,n>N$. Hence, $\{a_{n}\}_{n\in\mathbb{N}}$ is not Cauchy.
At this point, realising that a non-Cauchy sequence can yet have a convergent subsequence, I became stuck. I wonder whether my stronger result implies that the sequence I constructed cannot have a convergent subsequence.
I'm sorry to inform you that the method u've presented so far needs some modification to continue. The problem comes from the way you construct a sequence divergent to infinity.To show that why this type of construction could give us a sequence that has convergent subsequence, we assume that$(X,d)=\mathbb{R}$ and K unbounded but compact. Let $\{x_n\}$ be a sequence defined as follows : $$\forall n\in\mathbb{N},x_{2n}=2n,x_{2n-1}=0$$ It is clear that $\forall n\in \mathbb{N}, d(x_{n+1},x_n)=|x_{n+1}-x_n|=2n>n$. However it has a convergent subsequence $\{x_{2n-1}\}$.
So why such construction does not work well? In fact, if you choose "arbitrarily" a point to be $a_{n+1}$ such that $d(a_n,a_{n+1})>n$, then you have the risk that $a_{n+1}$ lives close to some point before ($a_1,a_3,a_{3k-1}$, e.t.c).Therefore, if you want to continue this method (that is, to construct a sequence divergent to infinity),you should specify the way you choose $\{a_n\}$'s to avoid the case mentioned above.