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I'm having difficulty understanding the following inequality as presented in Kenneth A Ross's Elementary Analysis: The Theory of Calculus $(2^{nd} Edition)$.

For a sequence $(s_n) \in \mathbb R$, let $v_N$ = $\mathrm{sup}\{s_n : n > N\}$.

$v_N$ = $\mathrm{sup}\{s_n : n > N\}$ $\implies$ $v_1 \ge v_2 \ge v_3 ... $

This makes half-sense to me. For $S = \{s_n : \forall n \in \mathbb N \} $, isn't $\mathrm{sup}\, S$ universal--ie, for $v_N$ = $\mathrm{sup}\{s_n : n > N\}$, shouldn't $v_1 = v_2 = v_3 = ...;$ $\,$?

Or is my line of reasoning incorrect because a general sequence does not necessarily converge to it's supremum? The way I'm trying to think of this is by visualizing a monotonic increasing sequence on a number line (bounded above). Even though you're taking $s_n \, \forall \, \mathrm n > \mathrm N$ , wouldn't you still hit the same supremum? Seeing as the supremum is defined as the least upper bound, how could the least upper bound "change"

Or am I misunderstanding things because I'm not considering how general this statement is?

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  • $\begingroup$ What if the sequence is decreasing? $\endgroup$
    – user1337
    Dec 13, 2014 at 22:22

1 Answer 1

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But you're 'losing' points in, $v_N = \mathrm{sup}\{s_n : n > N\}$, take for example $s_n = \dfrac{1}{n}$ . Then $v_N = \dfrac{1}{N+1}$ and $v_1 \neq v_2 \neq v_3 \dots$

There is also a theorem that states:

if $A,B$ are nonempty bounded sets of real numbers and $A \subseteq B$ then $\sup A \leq \sup B$

Proof:

Because they're bounded and nonempty we have $\sup A = a$, $\sup B = b$, clearly, $b$ is an upper bound for the elements of $A$, meaning if $c \in A$, $c \leq b$, now, suppose $a > b$, then $a -b > 0$,take $\varepsilon = a-b$. By properties of $\sup$, there is some $c \in A$ such that $a - \varepsilon < c$, but this means that $b < c$, a contradiction.

And from this it follows that $v_1 \geq v_2 \geq v_3 \dots$

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