Proof Explanation: Convergence implies boundedness

Question

Hello. The picture/proof is taken from Bartle. I'm having trouble understanding the proof. Does $|x_n| < 1+ |x|$ imply that all the elements of the sequence that are come after n>K will be less than this number?

Moreover, when we take the supremum, we're just looking for an upper bound right? Because if it so happens that the supremum is $1 + |x|$ then it is not necessarily the supremum of the sequence. However, if it is one of the elements from $ |x_1|, |x_2|,.....,|x_{k-1}|$, then the bound we get is also a supremum of the sequence?

Also, since this is just talking about boundedness, why not consider the case for a decreasing sequence? I feel like this proof is not tackling that case?

Please correct me if I'm wrong.

Thank you

Answer 1

Bartle's proof says $|x_n|< 1+|x|$ for $n\ge K$. As stated in the proof, it follows from $|x_n-x|<1$ for all $n>K$. Bartle's proof just says any convergent sequence has a bound. More precisely, if $\langle x_n\rangle$ is a convergent sequence then we can find $M>0$ such that $|x_n|< M$. Espesially, $\langle x_n\rangle$ has $-M$ as a lower bound.

I don't understand why you mention decreasing seqences. Decreasing sequences are not necessarily bounded; for example, $x_n=-n$.

Answer 2

1) Yes, all elements $x_n$, for $n\gt K$, will satisfy $\lvert x_n\rvert \lt 1+\lvert x \rvert $...

2) The sup $M$ will be a bound for the sequence (if it's one of $\lvert x_1 \rvert , \dots,\lvert x_{K-1} \rvert$ it will in fact be the least upper bound, or sup, of the sequence: you are correct.)

3) The proof applies to any convergent sequence of real numbers (even a decreasing sequence).