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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?

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

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  • $\begingroup$ 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! $\endgroup$ – Masha Apr 30 at 4:33
  • $\begingroup$ @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 $(y_n)$. $\endgroup$ – Theo Bendit Apr 30 at 4:37
  • $\begingroup$ 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$$ ? $\endgroup$ – Masha Apr 30 at 4:44
  • $\begingroup$ @Masha that's correct. $\endgroup$ – rubikscube09 Apr 30 at 5:04
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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.

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  • $\begingroup$ 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 $z_n =x_n + y_n \equiv 0$ $\endgroup$ – rubikscube09 Apr 30 at 5:05
  • $\begingroup$ $x_n$ and $y_n$ are both strictly increasing so this cannot be the case. $\endgroup$ – nilradical1 Apr 30 at 15:59
  • $\begingroup$ Right. My mistake. I thought you were referencing the general case. $\endgroup$ – rubikscube09 Apr 30 at 16:08

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