Do I need induction to prove that this sequence is monotonic?

[CoffeeNerd999]

TL;DR Summary

If i'm proving $y_n = \sup\{x_j | j \geq n\}$ is decreasing, can this be done without invoking mathematical induction?

I think the initial assumptions would allow me to prove this without induction.

Suppose $(x_n)$ is a real sequence that is bounded above. Define $$ y_n = \sup\{x_j | j \geq n\}.$$

Let $n \in \mathbb{N}$. Then for all $j \in \mathbb{N}$ such that $j \geq n + 1 > n$
$$ x_{j} \leq y_n.$$ So, $y_n$ is an upper bounded of $(x_j)_{j=n+1}^\infty$. By definition, $y_{n+1}$ is the least upper bound of $(x_j)_{j=n+1}^\infty$, so
$$y_{n+1} \leq y_n.$$ Since $n$ was chosen arbitrarily, this proves $y_n$ is monotone decreasing.

Am I correct that the using the supremum's definition makes using the induction principle suprifilous for this result?

 

[member 587159]

Your proof is correct and your observations as well. You don't need mathematical induction here. The following is the general situation that you applied: $$\emptyset \neq A \subseteq B \implies \sup A \leq \sup B$$

This is rather straightforward to prove (essentially the same proof you gave): If $\sup B = +\infty$, there is nothing to prove. So we may assume that $\sup B < +\infty$. Recall that $\sup B$ is by definition an upperbound for all elements in $B$, thus also for all elements of $A$ because $A \subseteq B$. By definition of supremum, we must have $\sup A \leq \sup B$ ($\sup A$ is the smallest upper bound for $A$ and $\sup B$ is an upper bound for $A$).

Since, $\{x_j: j \geq n\} \supseteq \{x_j: j \geq n+1\}$, this will allow you to conclude that $y_n \geq y_{n+1}$, which is what you want.