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Let $(x_n)$ and $(y_n)$ be strictly increasing sequences, and let $(z_n)$ be a sequence defined by $z_n = x_n + y_n$ for all $n \in \mathbb{N}$.

Prove that if $(z_n)$ is bounded above, then so are $(x_n)$ and $(y_n)$.

I do not know where to start with this problem. I know that $(z_n)$ being bounded above means there exists some $A \in \mathbb{N}$ such that $z_n < A$ for all $n \in \mathbb{N}$, therefore $x_n + y_n < A$ for all $n \in \mathbb{N}$. I don't see how this helps finding some $B \in \mathbb{N}$ such that $x_n < B$ (or $y_n < B$).

I have also tried proving the contrapositive but it did not get me anywhere.

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  • $\begingroup$ A 'reductio ad absurdum' seems to be a nice approach... $\endgroup$
    – Dr. Mathva
    Feb 28 '19 at 21:32
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    $\begingroup$ Hint: Note that $z_n > x_n + y_0$, so $x_n < z_n - y_0$. $\endgroup$ Feb 28 '19 at 21:32
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Assume that $(x_n)_n$ is not bounded from above. Then $x_n \to\infty$ so $$z_n = x_n + y_n > x_n + y_1 \to \infty$$ Hence $(z_n)_n$ is not bounded from above.

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Let $M$ be a bound for $(z_n)_n$. Then $M-y_0$ is a bound for $(x_n)_n$ (and similary, $M-x_0$ a bound for $(y_n)_n$): $$x_n=z_n-y_n\le z_n-y_0\le M-y_0. $$

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One has $x_n = z_n - y_n$. The sequence $(z_n)$ is bounded above, and the sequence $(y_n)$ is inreasing, so $(-y_n)$ is decreasing so it's also bounded above. So $(x_n)$ is the sum of two sequences that are bounded above, so it is bounded above.

You can do the same for $(y_n)$.

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