# Show that the if a sequence is bounded then show that the two sequences that make it up are bounded

I am trying to show given $$z_n = x_n + y_n$$ and $$(x_n)$$ and $$(y_n)$$ are both strictly increasing sequences, show that if $$(z_n)$$ is bounded above, then $$(x_n)$$ and $$(y_n)$$ is bounded above.

This is what I have so far: If $$(z_n)$$ is bounded above, then $$\exists M$$ such that $$z_n \leq M$$ and $$M$$ is an upper bound for $$(z_n)$$. Since $$z_n = x_n + y_n$$, then $$x_n + y_n \leq M$$.

So, this means that $$M$$ is an upper bound for $$x_n + y_n$$. However, how can I show that both $$(x_n)$$ and $$(y_n)$$ is bounded above?

## 2 Answers

Hint. Following on from what you have already done, if $$(y_n)$$ is increasing then $$x_n\le M-y_n\le M-y_1\ .$$

• Hi, I'm sorry I'm still not really sure where to go from here. Can you give me an other hint for this hint? Thank you! – Masha Apr 30 at 4:33
• @Masha So, $x_n \le M - y_1$ for all $n$, making $(x_n)$ bounded above by the constant $M - y_1$. Try doing the same for $(y_n)$. – Theo Bendit Apr 30 at 4:37
• That makes sense - thank you! So for $(y_n)$ if $(x_n)$ is increasing then $y_n \leq M - x_n \leq M - x_1$ and then it follows from the same logic as above that $(y_n)$ is bounded above by the constant $M - x_1$$? – Masha Apr 30 at 4:44 • @Masha that's correct. – rubikscube09 Apr 30 at 5:04 If either of $$x_n$$ or $$y_n$$ were unbounded, then it would not be possible to put a bound on $$z_n = x_n + y_n$$. Thus they must be bounded. • If$x_n = -y_n$for all$n$then it is possible that$x_n$is bounded below but not bounded above and similarly$y_n$is bounded above and not below, and$z_n =x_n + y_n \equiv 0$– rubikscube09 Apr 30 at 5:05 •$x_n$and$y_n\$ are both strictly increasing so this cannot be the case. – nilradical1 Apr 30 at 15:59
• Right. My mistake. I thought you were referencing the general case. – rubikscube09 Apr 30 at 16:08