Every Cauchy sequence is bounded. Question about a detail in the proof. To prove that every Cauchy sequence is bounded, we say that after some $k$ all $x_n$ are contained in a ball of given radius $\epsilon$ for $n \geq k$. We then say all $x_n$ with $1 \leq n \leq k-1$ are contained in a ball with radius given as the maximum distance between any two $x_n$ with $1 \leq n \leq k-1$. We make an argument that includes assuming this distance will be finite. 
How do we know the distance will be finite? Can I not have a Cauchy sequence that comes in from infinity?
 A: Suppose we are in $\mathbb{R}$. It will be a natural extension to other metric spaces.
Proof: Let $\{a_j\}$ be a Cauchy sequence. We need to show $\exists M>0, M\in \mathbb{R}, s.t. |a_j|\leq M$, where $j=1,2,3,,,$. (Recall this is exactly the definition of boundedness.)
Well, at this moment, you might be tempted to let $M=max\{|a_1|,|a_2|,|a_3|,...,\}$. This is wrong, because we are not sure whether this maximum exists or not. It contains infinite terms. This is the difference between infinity and finiteness. For example, if $a_j=j,j=1,2,3,,,$, then you can easily see there is no maximum. But if this sequence only contains finite terms, then it is bounded.
Thus, we need to improve this approach. But anyway, the above idea is broadly correct. The correct proof is below:
Because $\{a_j\}$ is Cauchy, thus according to definition, $\forall \epsilon >0, \exists N\in \mathbb{N}$ such that whenever $j,k>N$, we have $|a_j-a_k|<\epsilon$.
Now for $\epsilon =1>0,\exists N\in \mathbb{N} $, call it $N_1,$ s.t. whenever $j,k>N_1$, we have $|a_j-a_k|<1$. Specifically, let $k=N_1+1>N_1,j>N_1$, then we have $|a_j-a_{N_1+1}|<1$.
This implies $|a_j|\leq|a_j-a_{N_1+1}|+|a_{N_1+1}|<1+|a_{N_1+1}|$. Notice that this line holds only when $j>N_1$.
Now $max\{|a_1|,|a_2|,|a_3|,...,|a_{N_1}|,1+|a_{N_1+1}|\}=M\in \mathbb{R}, M>0$ and $|a_j|\leq M$ for all $j$.
This $M$ is exactly what we need for the proof. Because when $j=1,2,3,...,N_1$,$|a_j|\leq M$. When $j>N_1$, then $|a_j|\leq 1+|a_{N_1+1}|\leq M$.
A: Something to think about: If the sum is coming in from $\infty$, what is $x_1$?, the first term of the sequence?
