# Let $z_n = x_n + y_n$, with $(x_n)$ and $(y_n)$ strictly increasing. Prove that if $(z_n)$ is bounded above, then so are $(x_n)$ and $(y_n)$.

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.

• A 'reductio ad absurdum' seems to be a nice approach... Commented Feb 28, 2019 at 21:32
• Hint: Note that $z_n > x_n + y_0$, so $x_n < z_n - y_0$. Commented Feb 28, 2019 at 21:32

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.
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 $$(y_n)_n$$): $$x_n=z_n-y_n\le z_n-y_0\le M-y_0.$$
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 $$(y_n)$$.