Proof: A convergent Sequence is bounded A part of the proof says that if $n\le N$, then, the sequence 
$x_n \le \max\{|x_1|,|x_2|,....,|x_{N-1}|\}$.
I'm not capturing the intuition of the above. Even more perplexing is, if it the case where the sequence is monotonic decreasing, then for every $n\le N$, it is obvious that $x_n$ will not be $\le \max\{|x_1|,|x_2|,....,|x_{N-1}|\}$.
I greatly appreciate some input.
 A: Well, $|x_n| \le \max\{|x_1|,\ldots, |x_{N-1}|\}$ for all $n \color{red}{<} N$ because if $n < N$, $|x_n|$ is one of the numbers $|x_1|,\ldots, |x_{N-1}|$, and some number in this list is the largest, i.e., $\max\{|x_1|,\ldots, |x_{N-1}|\}$.
A: I think it is more useful to just think of this in an arbitrary metric space than to worry about properties of the real numbers such as order (i.e. monotone sequences), because it's not actually relevant to the fact that a convergent sequence is bounded.
If $\lim_{n\to\infty} x_n=x$, then for all $\varepsilon>0$, there exists $N$ such that $n\geqslant N$ implies $d(x,x_n)<\varepsilon$.
So take $\varepsilon=1$. Choose $N$ so that $n\geqslant N$ implies $d(x,x_n)<1$. Then for any $n$, 
$$d(x,x_n) \leqslant \max\left\{1, \max_{1\leqslant m\leqslant N}\{d(x,x_m)\}\right\}. $$
Hence, $\{x_n\}$ is bounded.
For a more intuitive, but less formalized argument; if the sequence is convergent, then we can draw an open ball around the limit that contains all but finitely many points of the sequence. So, draw an open ball around those finitely many points, and take the union of those two balls. Draw an even bigger ball that contains that union. Then you have a ball that contains the entire sequence.
For yet another argument: suppose $\{x_n\}$ isn't bounded. Then no open ball centered at $x$ contains all points of the sequence. So for each $k$, we can find a point $x_{n_k}$ such that $d(x,x_{n_k})>k$. Take $\varepsilon=1$. Then for any $N$, we have a positive integer $n_k$ such that $d(x,x_{n_k})>k>1$. Hence, $x_n$ does not converge to $x$, a contradiction.
A: I see what's going on.
My lecturer could have done a better job by stating, instead of 
"If n < N", "for all n in the set of Natural number"
Since 1+|L| is a bound for x_n for which n>=N, the bound 1+|L| might not necessarily be the bound for x_n since we have not taken into account yet x_n for which n < N.
A: If $\{a_n\}$ be a convergent then for any $\epsilon>0$: then there exist a positive integer $n_0$ such that 
$$
|a_n-l|\le \epsilon\:\text{  for all }\:n> n_0
$$
and this implies that $|a_n|\le\epsilon +l$.
Now let $M=\{|a_1|,|a_2|,|a_3|,\ldots,|a_n| ,|\epsilon+l|\}$: then $|a_n|\le M$ is bounded for all $n\in\Bbb N$ and hence sequence $\{a_n\}$ is bounded.
