I'm currently studying, or revising, Walter Rudin's Principles of Mathematical Analysis, 3rd eition, and I'm stuck with the proof of Theorem 3.17.

Here's the statement of the Theorem:

Let $\{s_n\}$ be a sequence of real numbers, let $E$ be the set of all subsequential limits of $\{s_n\}$ in the extended real number system, and let $$s^* \colon= \sup E.$$ That is, let $s^*$ be the upper limit (or limit superior) of the sequence $\{s_n\}$. Then

(a) $s^* \in E$.

(b) If $x > s^*$, there is an integer $N$ such that $n \geq N$ implies $s_n < x$.

Moreover, $s^*$ is the only number with the properties (a) and (b).

I've figured out that if $s^* = +\infty$, then the set $E$ of subsequential limits is unbounded above in $\mathbb{R}$; so for every positive integer $N$ there is an element $x^{(N)}$ of $E$ such that $x^{(N)} > N$. Now since $x^{(N)}$ is the limit of a subsequence, say $\{s_{\phi_N(n)}\}$, of the sequence $\{s_n\}$, so there is a natural number $n_N$ such that $s_{n_N} > N$. So the subsequence $\{s_{n_N}\}$ diverges to $+\infty$. Am I right?

If $s^*$ is real, then the set $E$ is bounded above, but how to convince ourselves that in this case the set $E$ is NON-EMPTY in the first place?

And, if $s^* = -\infty$, then the set $E$ has no real number as an element; so that the only possible element of $E$ is $-\infty$. Now how to go about showing that this IS actually the case?

Finally, how to show that if $x > s^*$ and if $s_n \geq x$ for infinitely many $n$, then there exists an element $y$ in $E$ such that $y > s^*$?


1 Answer 1

  1. Yes. You should find a divergent subsequence so that $+\infty \in E$.
  2. We can show that $E$ cannot be empty. If $\{s_n\}$ is unbounded above or below, then $+\infty \in E$ or $-\infty \in E$ for reasons similar to 1. If $\{s_n\}$ is bounded, then by Bolzanno Weierstrass there must be a convergent subsequence, so $E$ again is nonempty.
  3. That simply follows from the definition. If $s^\ast = -\infty$ is the supremum of $E$, then the existence of any real number in $E$ would make $s^\ast$ not an upper bound. Contradiction.
  4. Take $\{s_{k_n}\}$ be the subsequence of all elements at least $x$. By 2., the set of subsequence limit points of $\{s_{k_n}\}$, call it $A \subset E$ is nonempty. Let $y \in A$. Since $s_{k_n} \geq x$ for all $n$, $y$ cannot be less than $x$. Thus $y \geq x > s^\ast$.

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