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CoffeeNerd999

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- 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?

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?