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Here're Definitions 3.15 and 3.16 and Theorem 3.17 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition.

Definition 3.15:

Let $\{s_n \}$ be a sequence of real numbers with the following property: For every real $M$ there is an integer $N$ such that $n \geq N$ implies $s_n \geq M$. We then write $$s_n \rightarrow +\infty.$$ Similarly, if for every real $M$ there is an integer $N$ such that $n \geq N$ implies $s_n \leq M$, we write $$s_n \rightarrow -\infty.$$

Definition 3.16:

Let $\{ s_n \}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} \rightarrow x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits as defined in Definition 3.5, plus possibly the numbers $+\infty$, $-\infty$.

We now recall Definitions 1.8 and 1.23 and put $$s^* = \sup E,$$ $$s_* = \inf E.$$ The numbers $s^*$, $s_*$ are called the upper and lower limits of $\{ s_n \}$; we use the notation $$\lim_{n\to\infty} \sup s_n = s^*, \ \ \ \lim_{n\to \infty} \inf s_n = s_*.$$

Now,

Definition 3.5:

Given a sequence $\{ p_n \}$, consider a sequence $\{n_k\}$ of positive integers, such that $n_1 < n_2 < n_3 < \cdots$. Then the sequence $\{p_{n_k} \}$ is called a subsequence of $\{p_n \}$. If $\{p_{n_k}\}$ converges, its limit is called a subsequential limit of $\{p_n \}$.

Definition 1.8:

Suppose $S$ is an ordered set, $E \subset S$, and $E$ is bounded above. Suppose there exists an $\alpha \in S$ with the following properties:

(i) $\alpha$ is an upper bound of $E$.

(ii) If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $E$.

Then $\alpha$ is called the least upper bound of $E$ or the supremum of $E$, and we write $$\alpha = \sup E.$$

The greatest lower bound, or infimum, of a set $E$ which is bounded below is defined in the same manner: The statement $$\alpha = \inf E$$ means that $\alpha$ is a lower bound of $E$ and that no $\beta$ with $\beta > \alpha$ is a lower bound of $E$.

Definition 1.23:

The extended real number system consists of the real field $R$ and two symbols, $+\infty$ and $-\infty$. We preserve the original order in $R$, and define $$-\infty < x < +\infty$$ for every $x \in R$.

It is then clear that $+\infty$ is an upper bound of every subset of the extended real number system, and that every nonempty subset has a least upper bound. If, for example, $E$ is a nonempty set of real numbers which is not bounded above in $R$, then $\sup E = +\infty$ in the extended real number system.

Exactly the same remarks apply to lower bounds.

Finally, here's Theorem 3.17.

Let $\{s_n \}$ be a sequence of real numbers. Let $E$ and $S^*$ have the same meaning as in Definition 3.16. Then $s^*$ has the following two properties:

(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).

Of course, an analogous result is true for $s_*$.

And, here's Rudin's proof:

(a) If $s^* = +\infty$, then $E$ is not bounded above; hence $\{s_n\}$ is not bounded above, and there is a subsequence $\{s_{n_k}\}$ such that $s_{n_k} \rightarrow +\infty$.

If $s^*$ is real, then $E$ is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28. [I'll state these theorems after the proof. ]

If $s^* = -\infty$, then $E$ contains only one element, namely $-\infty$, and there is no subsequential limit. Hence, for any real $M$, $s_n > M$ for at most a finite number of values of $n$, so that $s_n \rightarrow -\infty$.

This establishes (a) in all cases.

(b) Suppose there is a number $x > s^*$ such that $s_n \geq x$ for infinitely many values of $n$. In that case, there is a number $y \in E$ such that $y \geq x > s^*$, contradicting the definition of $s^*$.

Thus $s^*$ satisfies both (a) and (b).

To show the uniqueness, suppose there are two numbers, $p$ and $q$, which satisfy (a) and (b), and suppose $p < q$. Choose $x$ such that $p < x < q$. Since $p$ satisfies (b), we have $s_n < x$ for $n \geq N$. But then $q$ cannot satisfy (a).

Now for Theorem 2.28:

Let $E$ be a nonempty set of real numbers which is bounded above. Let $y = \sup E$. Then $y \in \overline{E}$. Hence $y \in E$ if $E$ is closed.

And, Theorem 3.7:

The subsequential limits of a sequence $\{p_n\}$ in a metric space $X$ form a closed subset of $X$.

My reason for copying so many definitions and theorem from Rudin's book into my question is to keep my post as self-contained as possible.

Now here's my attempt at filling in the details in the proof of Theorem 3.17.

(a) If $s^* = +\infty$, then the set $E$ is not bounded above in $\mathbb{R}$. We now find a subsequence of $\{s_n \}$ which diverges to $+\infty$.

As $E$ is not bounded above in $\mathbb{R}$, so there is an element $x_1 \in E$ such that $x_1 > 1$. If $x_1 = +\infty$, then we are done. Otherwise, $x_1$ is a subsequential limit of $\{s_n\}$. Thus there is a strictly increasing function $\varphi_1 \colon \mathbb{N} \to \mathbb{N}$ such that $$x_1 = \lim_{n\to\infty} s_{\varphi_1(n)}.$$ So, for any real number $\varepsilon > 0$, we can find a natural number $N_1$ such that $n > N_1$ implies that $$ \left\vert s_{\varphi_1(n)} - x_1 \right\vert < \varepsilon.$$ Let's take $\varepsilon$ such that $$ 0 < \varepsilon < \frac{x_1 - 1}{2}.$$ Then we can conclude that $$s_{\varphi_1(N_1 + 1)} > 1.$$ Let $$n_1 \colon= \varphi_1(N_1+1).$$ Then $$s_{n_1} > 1.$$ Now as $E$ is not bounded above in $\mathbb{R}$, there is an element $x_2 \in E$ such that $x_2 > 2$. If $x_2 = +\infty$, then we are done. Otherwise, $x_2$ is a subsequential limit of $\{s_n\}$. That is, there is a strictly increasing function $\varphi_2 \colon \mathbb{N} \to \mathbb{N}$ such that $$x_2 = \lim_{n\to\infty} s_{\varphi_2(n)}.$$ So, for any real number $\varepsilon > 0$, we can find a natural number $N_2$ such that $n > N_2$ implies that $$ \left\vert s_{\varphi_2(n)} - x_2 \right\vert < \varepsilon. $$ Thus, if we take $\varepsilon$ such that $$0 < \varepsilon < \frac{x_2 - 2}{2},$$ then we can conclude that $$s_{\varphi_2(N_2 + 1)} > 2.$$ Let's put $$n_2 \colon= \max \left\{ \varphi_2( n_1 +1 ), \varphi_2(N_2+1) \right\} .$$ We note that since $\varphi_2 \colon \mathbb{N} \to \mathbb{N}$ is a strictly increasing function, we can show that $$n \leq \varphi_2(n)$$ for all $n \in \mathbb{N}$. So we have $$n_2 \geq \varphi_2(n_1 + 1) \geq n_1 + 1 > n_1.$$

Now we can find an element $x_3 \in E$ such that $x_3 > 3$. If $x_3 = +\infty$, then we are done. Otherwise, $x_3$ is a subsequential limit of $\{s_n \}$. That is, there is a strictly increasing function $\varphi_3 \colon \mathbb{N} \to \mathbb{N}$ such that $$x_3 = \lim_{n \to \infty} s_{\varphi_3(n)}.$$ So, for any real number $\varepsilon > 0$, we can find a natural number $N_3$ such that $n > N_3$ implies that $$\left\vert s_{\varphi_3(n)} - x_3 \right\vert < \varepsilon.$$ So, if we take $\varepsilon$ such that $$0 < \varepsilon < \frac{x_3-3}{3},$$ then we can conclude that $$s_{\varphi_3(N_3+1)} > 3.$$ Let's take $$n_3 \colon= \max \left\{ \varphi_3 ( n_2+1), \varphi_3 (N_3 + 1) \right\}.$$ Since $\varphi_3 \colon \mathbb{N} \to \mathbb{N}$ is a strictly increasing function, we can show that $$ n \leq \varphi_3(n) $$ for all $n \in \mathbb{N}$. So $$n_3 \geq \varphi_3(n_2 + 1) \geq n_2 + 1 > n_2.$$

In this way, we obtain a sequence $\{n_k\}$ of natural numbers such that $n_1 < n_2 < n_3 < \cdots$ and $$s_{n_k} > k$$ for all $k \in \mathbb{N}$. Now for any real number $M$, there is a natural number $K$ such that $K > M$. So, for all $k \in \mathbb{N}$ such that $k > K$, we have $$s_{n_k} > k > K > M;$$ that is, $s_{n_k} > M$ for all $k \in \mathbb{N}$ such that $k > K$. Therefore, we can conclude that $$s_{n_k} \rightarrow +\infty,$$ which shows that $E$ contains $+\infty$.

If $s^* \in \mathbb{R}$, then $E$ is bounded above in $\mathbb{R}$ and $s^* = \sup E$. We show that some subsequence of $\{s_n \}$ converges to $s^*$. There exists an element $x_1 \in E$ such that $s^*-1 < x_1 \leq s^*$. If $x_1 = s^*$, then we are done. Otherwise we have $s^*-1 < x_1 < s^*$. Now $x_1$ is a subsequential limit of $\{s_n\}$. So there is a strictly increasing function $\varphi_1 \colon \mathbb{N} \to \mathbb{N}$ such that $$x_1 = \lim_{n\to\infty} s_{\varphi_1(n)}.$$ Thus, for any real number $\varepsilon > 0$ we can find a natural number $N_1$ such that $n > N_1$ implies $$\left\vert s_{\varphi_1(n)} - x_1 \right\vert < \varepsilon. $$ By taking $\varepsilon$ such that $$ 0 < \varepsilon < \frac 1 3 \min \{ s^* - x_1, \ x_1 - s^* + 1 \},$$ we can conclude that $$s^* -1 < s_{\varphi_1(N_1 + 1) } < s^*.$$ Let's take $$n_1 \colon= \varphi_1(N_1 + 1).$$

As $s^* = \sup E$ in $\mathbb{R}$, there is an element $x_2 \in E$ such that $s^* - \frac 1 2 < x_2 \leq s^*$. If $x_2 = s^*$, then we are done. Otherwise we have $$s^*- \frac 1 2 < x_2 < s^*.$$ Now $x_2$ is a subsequential limit of $\{s_n\}$. So there is a strictly increasing function $\varphi_2 \colon \mathbb{N} \to \mathbb{N}$ such that $$x_2 = \lim_{n\to\infty} s_{\varphi_2(n)}.$$ So, given a real number $\varepsilon > 0$, we can find a natural number $N_2$ such that $n>N_2$ implies that $$\left\vert s_{\varphi_2(n)} - x_2 \right\vert < \varepsilon.$$ Thus, by taking $\varepsilon$ such that $$0< \varepsilon < \frac 1 3 \min \left\{ s^* - x_2 , \ x_2 - s^* + \frac 1 2 \right\},$$ we can conclude that $$s^* - \frac 1 2 < s_{\varphi_2(N_2 + 1)} < s^*.$$ Let's take $$n_2 \colon= \max \left\{ \varphi_2(n_1 + 1), \varphi_2(N_2+ 1) \right\}.$$ Then $n_2 > n_1$ and $$s^* - \frac 1 2 < s_{n_2} < s^*.$$

Now as $s^* = \sup E$ in $\mathbb{R}$, we can find an element $x_3 \in E$ such that $$s^* -\frac 1 3 < x_3 \leq s^*.$$ If this $x_3 = s^*$, then we are done. Otherwise we have $$s^* - \frac 1 3 < x_3 < s^*.$$ Moreover, $x_3$ is a subsequential limit of $\{s_n\}$. So there is a strictly increasing function $\varphi_3 \colon \mathbb{N} \to \mathbb{N}$ such that $$x_2 = \lim_{n\to\infty} s_{\varphi_3(n)}.$$ Thus, given a real number $\varepsilon > 0$, we can find a natural number $N_3$ such that $n > N_3$ implies $$\left\vert s_{\varphi_3(n)} - x_3 \right\vert < \varepsilon.$$ Thus, by taking $\varepsilon$ such that $$0< \varepsilon < \frac 1 3 \min \left\{ s^* - x_3 , \ x_3 - s^* + \frac 1 3 \right\},$$ we can conclude that $$s^* - \frac 1 3 < s_{\varphi_3(N_3 + 1)} < s^*.$$ Let's take $$n_3 \colon= \max \left\{ \varphi_3(n_2 + 1), \varphi_3(N_3+1) \right\}.$$ Then $n_3 > n_2$ and $$s^* - \frac 1 3 < s_{n_3} < s^*.$$

Continuing in this way, we obtain a sequence $\{n_k \}$ of natural numbers such that $n_1 < n_2 < n_3 < \cdots$ and $$s^* - \frac 1 k < s_{n_k} < s^*$$ for all $k \in \mathbb{N}$. From here, we can show that the subsequence $\{s_{n_k}\}$ of $\{s_n\}$ converges to $s^*$.

If $s^* = -\infty$, then there is no subsequential limit of $\{s_n\}$. Moreover, $\{s_n \}$ is bounded above in $\mathbb{R}$, for otherwise there would be a subsequence of $\{s_n \}$ which would diverge to $+\infty$.

Let $M$ be any real number. If we were to have $s_n \geq M$ for infinitely many values of $n \in \mathbb{N}$, then we could find a subsequence of $\{s_n\}$ which would be bounded and hence have a convergent subsequence (by Theorem 3.6 (b) in Rudin), which would in turn be a convergent subsequence of $\{s_n\}$ whose limit $x$ would satisfy $x \in E$ and $x > s^*$, a contradiction.

So we must have $s_n \geq M$ for at most a finite number of values of $n$. Thus, there is a natural number $N$ such that $n > N$ implies that $s_n < M$. Since $M$ was an arbitrary real number, we can conclude that $$s_n \rightarrow -\infty, $$ which shows that $-\infty \in E$.

Thus we have shown that $s^* \in E$ in all possible cases.

Now part (b):

We assume the contrary. That is, we assume that $x > s^*$ and that there are infinitely many $n$ such that $s_n \geq x$. Since $s^* < x$, $s^*$ cannot be $+\infty$. Moreover, there is a subsequence $\{s_{\varphi(n)}\}$ of $\{s_n\}$ such that $$s_{\varphi(n)} \geq x$$ for all $n \in \mathbb{N}$, where $\varphi \colon \mathbb{N} \to \mathbb{N}$ is a strictly increasing function.

If the subsequence $\{s_{\varphi(n)}\}$ is not bounded above, then some subsequence of $\{s_{\varphi(n)}\}$ --- which would in turn be a subsequence of $\{s_n\}$ --- would diverge to $+\infty$, showing that $+\infty \geq x > s^*$ and $+\infty \in E$, a contradiction.

On the other hand, if the subsequence $\{s_{\varphi(n)}\}$ is bounded above, then it would be a bounded sequence in $\mathbb{R}$ and so would have a convergent subsequence (by Theorem 3.6(b) in Rudin), which would in turn be a convergent subsequence of $\{s_n\}$, whose limit $s$ would satisfy $s \in E$ and $s \geq x > s^*$, which is a contradiction to the fact that $s^*$ is the supremum of $E$.

To show that $s^*$ is the only number with the properties (a) and (b), suppose that there are two numbers $p$ and $q$ which satisfiy properties (a) and (b) of Theorem 3.17 in Rudin, and suppose that $p < q$. Let's choose a real number $x$ such that $p < x < q$. Then by property (b), there is an integer $N$ such that $s_n < x$ for all $n \in \mathbb{N}$ such that $n \geq N$. Let us chooes a real number $\varepsilon$ such that $$0 < \varepsilon < \frac{ q-x}{3}.$$ Then there are at most a finite number of terms of $\{s_n\}$ in the neighborhood $(q-\varepsilon, q+ \varepsilon)$. So no subsequence of $\{s_n \}$ can converge to $q$ (or diverge to $+\infty$ if $q = +\infty$), showing that $q \not\in E$, which is a contradiction to the property (a) for $q$.

Based on how I have elaborated Rudin's proof in the last few paragraphs, have I been able to understand this proof correctly?

If not, where am I going wrong (or falling short)?

What is lacking in my reasoning, apart from brevity of course?

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  • 1
    $\begingroup$ Full marks for longevity. This breaks all the records. $\endgroup$ – user328032 Apr 25 '16 at 8:45
  • $\begingroup$ @Benedict yes, you're right, but it was partly intentionally lengthy. I wanted to make sure I understood the logic correctly. $\endgroup$ – Saaqib Mahmood Apr 25 '16 at 9:09
  • $\begingroup$ Why there is a finite number of terms of $\{s_n\}$ in the $(q-\varepsilon, q+\varepsilon)$? $\endgroup$ – AnalyticHarmony May 25 '17 at 12:35
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Your proof of 3.17 makes many assumptions that are not obvious to me.

Here is my proof of it for anyone who stumbles across this post and is similarly confused about 3.17.


3.17(a) case 1:

$s^* = +\infty$ imples that $E$ is not bounded above (suppose not, then we would have that sup $E < +\infty$).


$E$ is not bounded above implies that $\{s_n\}$ is not bounded above. To see why, suppose that $E$ is not bounded above, but $\{s_n\}$ is. Let $x$ = sup $\{s_n\}$. Since $E$ is not bounded above, there exists some $y \in E$ such that $y > x$. Thus, for some $\{s_{n_k}\}$, for all $\epsilon > 0$, there is some $N$ such that for all $n_k \ge N$, we have that $d(y, s_{n_k}) < \epsilon$.

Specifically, set $\epsilon = d(y, s)$. Then there exists an $s_{n_k}$ such that $d(y, s_{n_k}) < \epsilon$, which implies that $d(y, s_{n_k}) < d(y, x)$. We know that $y > x > s_{n_k}$ (by assumption and by the definition of $x$).

Therefore:

$d(y, s_{n_k}) < d(y, x)$

$\Rightarrow y - s_{n_k} < y - x$

$\Rightarrow s_{n_k} > x$, which contradicts the fact that $x =$ sup $\{s_n\}$.

Hence, we can conclude that $\{s_n\}$ is not bounded above!


We now construct a subsequence of $\{s_n\}$ as follows:

Let $s_{n_k} > k$ for all $k \in \mathbb{Z}^+$, with $n_1 < n_2 < n_3 < \dots$.

Suppose that we could not construct such a subsequence. If there were no $n_1$ such that $s_{n_1} \ge 1$, then $\{s_n\}$ would be bounded above (by 1), a contradiction. If there were no $n_k > n_i$ for $i \in \{1, \dots, k-1\}$ such that $s_{n_k} \ge k$, then we would have that $s_n < k$ for all $n > n_{k-1}$.

Thus, $\{s_n\}$ would be bounded by max($\{s_1, s_2, \dots, s_{n-k}\},k$), which contradicts the fact that $\{s_n\}$ is not bounded.

But this subsequence guarantees that for all $M$, there exists an $M$ such that there is some $N$ such that for all $n_k \ge N$, $s_{n_k} \ge M$. Hence $s_{n_k} \rightarrow +\infty$. Thus $+\infty \in E$.


3.17(a) case 2:

Now suppose $s^* \in \mathbb{R}$. Then $E$ is bounded above (by $s^*$). We now prove that this implies that at least one subsequential limit exists.

Suppose not. That is, suppose that $E$ were bounded above by $s^*$ and no subsequential limit existed. If no such limit existed, then there would be no $s \in \mathbb{R}$ such that $s \in E$. Hence, $E$ could only be four possible things: $\{+\infty, -\infty\}, \{+\infty\}, \{-\infty\}, \emptyset$. But we know that $+\infty \not \in E$, because if it were in $E$, that would contradict the fact that $s^* =$ sup $E$ is in $\mathbb{R}$.

Now suppose that either $E = \{-\infty\}$ or $E = \emptyset$. Then we would have that for any $y \in \mathbb{R}$ such that $y < s^*$, for all $x \in E$, $x < y$ (note this is obviously true when $E$ consists solely of $-\infty$ and is vacuously true when $E$ is the empty set). Hence we would have an upper bound of $E$ that is less than $s^*$ which contradicts the fact that $s^* = $ sup $E$. Thus, there is a subsequential limit of $\{s_n\}$.

By 3.7, we know that the subsequential limits of any sequence in metric space $X$ form a closed subset of $X$.

By 2.28, we know that for any non-empty subset $E$ of the real numbers that is bounded above, sup $E \in \overline{E}$.

Hence $s^* \in E$.


3.17(a) case 3:

Suppose that $-\infty =$ sup E. Then we know that $+\infty \not \in E$, and that $x \not \in E$ for all $x \in \mathbb{R}$.

We will now show that for every $M$, there exists an $N$ such that for all $n \ge N$, $s_n \le M$. This suffices to show that $s_n \rightarrow -\infty$, and hence that $-\infty \in E$.

Suppose not. That is, suppose that there is some $M$ such that for all $N$, there is some $n \ge N$ such that $s_n > M$. This implies that $s_n > M$ for infinitely many values of $s_n$. Let us consider the subsequence $\{s_{n_k}\}$ consisting of these values.

We know that $\{s_{n_k}\}$ is bounded below by $M$. There are two possibilities:

(I) $\{s_{n_k}\}$ is not bounded above. Hence, by the argument we made in 3.17(a) case 1, $+\infty \in E$, which contradicts the fact that $-\infty = s^* = $ sup $E$.

(II) $\{s_{n_k}\}$ is bounded above. Then $\{s_{n_k}\}$ is bounded below by $M$ and above by $\beta$, for some $\beta$. Hence, $\{s_{n_k}\}$ is bounded. But by 3.6(b) every bounded sequence in $\mathbb{R}$ contains a convergent subsequence. Hence, some subsequence of $\{s_n\}$ converges to some point $p \in \mathbb{R}$, which contradicts the fact that $-\infty = s^* =$ sup $E$.

Thus, for all $M$, $s_n > M$ for only finitely many values of $s_n$. Thus $s_n \rightarrow -\infty$. Therefore, $s^* \in E$.

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