Proof of $ \sup\limits_{n\in\mathbb{N}} (x_n+y_n) \le\sup x_n+\sup y_n$

Question

For sequence $(a_n)_{n\in \Bbb N}$ we define: $\sup_{n\in \Bbb N} a_n=\{a_{n}:n\in\Bbb N\}$: prove that for each pair of sequences $(x_{n})_{n\in \Bbb N}, (y_{n})_{n\in \Bbb N}$ which are bounded from above $$ \sup_{n\in \Bbb N} (x_n+y_n) \le \sup x_n+\sup y_n $$ and point out such pair of sequences, for which presented inequality is "$<$".

My attempt: let $a=\sup x_n$, and $b=\sup y_n$ and define the addition sequences as: $(a_n)+(b_n):=(a_n+b_n)$

1. When quantity of elements $(a_n)$ different from $(y_n)$:

   • $\sup(x_n+y_n)=\max\{a,b\},\text{ and }\sup x_n+\sup y_n=a+b$, thus
     $\max\{a,b\}\le a+b$ for all $a,b$.

2. When quantity of elements $(a_n)$ equals $(y_n)$:

   • $\sup(x_n+y_n)=a+b=\sup x_n+\sup y_n$

Does it makes sense?

Answer 1

Note that by definition supremum is the least upper bound.

Therefore it suffices to show that, $$ (x_n+y_n) \le \sup_{n\in \Bbb N}x_n+\sup_{n\in \Bbb N} y_n$$ because that makes $$\sup x_n+ \sup y_n$$ an upper bound for $\{ (x_n+y_n)\}$, $n\in \Bbb N$

Since $$ x_n \le \sup_{n\in \Bbb N} \{x_n\}$$ and $$ y_n \le \sup_{n\in \Bbb N} \{y_n\}$$ We get $$x_n+y_n \le \sup \{x_n\} +\sup \{y_n\}$$

Thus the least upper bound is less than or equal to this particular upper bound.

Answer 2

Here is a start on how one could proceed with proving this: Let $\varepsilon>0$ be some arbitrary positive number. Pick $N$ such that $$\sup(x_n+y_n)-\varepsilon<x_N+y_N$$

(why can we find such an $N$?). What upper bound can you place on the right-hand side?

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Edit: This is only applicable if we beforehand know that $\sup(x_n+y_n)$ is finite.

Answer 3

Let $a=\sup_{n\in\mathbb{N}}x_n$ and $b=\sup_{n\in\mathbb{N}}y_n$. Then, for every $n$, $$ x_n\le a,\qquad y_n\le b $$ and therefore $x_n+y_n\le a+b$, giving $$ \sup_{n\in\mathbb{N}}(x_n+y_n)\le a+b $$ because the supremum is at most equal to any upper bound.

It's generally false that $\sup_{n\in\mathbb{N}}(x_n+y_n)=\max\{a,b\}$. Take two constant sequences $x_n=1$ and $y_n=2$. Then $\sup_{n\in\mathbb{N}}=3\ne\max\{1,2\}$.

How do we find two sequences where the strict inequality holds?

Take $x_n=1$ when $n$ is even and $x_n=-1$ if $n$ is odd. Define $y_n=-x_n$.