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About Uniqueness Proofs

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My understanding of existence and uniqueness results is that the former condition asserts the cardinality of the "solution set" is positive, whereas the latter implies it is bounded above by $1$. It thus appears that both parts are integral. I can't help but notice, however, that sometimes the proofs resemble each other so much that one seems redundant, and I'm not sure if just one is sufficient. Consider Theorem 3.17 in Rudin's PMA, for which I gave a different proof than Rudin (I apologize in advance that I'm not the most adept at formatting, as a first-time poster on MSE):

Theorem 3.17: $\limsup_{n \to \infty} s_n$ is the only number (a) $s^* \in L$ such that (b) if $x > s^*$, then there exists an $N$ such that $n \geq N \implies s_n < x$.

My proof:

Let $s^* \in E$ be as described. It is clear that $s^* \leq \limsup_{n \to \infty} s_n$. Suppose for contradiction that $s^* < \limsup_{n \to \infty} s_n$. Then $s^*$ is not an upper bound of $L$, so there exists a $t \in E$ such that $s^* < t$. If $t = \infty$, then $L$, hence $(s_n)$, is not bounded above. But consider $s^* + 1$, which has associated an $N$ such that $n \geq N \implies s_n < s^* + 1$. Then $\{s_N, s_{N + 1}, \dots\}$ is bounded above, and so is $\{s_1, \dots, s_{N - 1}\}$, being finite, which yields a contradiction. Now suppose that $t \in \mathbb{R}$. Consider $m = \frac{s^* + t}{2}$, which furnishes an $N$ such that $n \geq N \implies s_n < m$. But there must exist an $M \geq N$ such that $t - s_M < t - m \implies s_M > m$, for otherwise all $m \geq N$ satisfies $|t - s_M| \geq t - s_M \geq t - m$, contrary to that $t$ is a sequential limit. Finally, it is impossible to have $t = -\infty$, as there is nothing smaller in the extended reals. The three contradictions conclude the proof.

Rudin's:

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 $s_n \to +\infty$.

If $s^*$ is real, then $E$ is bounded above, and at least one subsequential limit exists, so that existence follows from Theorems 3.7 and 2.28 (set of subsequential limits is closed, supremum in closed set).

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 $s_n \to -\infty$.

Hence $s^* \in E$.

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

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).

In this specific case, my concern arises mainly from that Rudin used the closeness property, while I haven't. But that the supremum is well-defined and uniquely determined by the conditions makes me wonder if it is actually necessary.

I wonder if there is an analog of, say, Schröder–Bernstein for this sort of situation, so that proving a slightly weaker statement suffices for existence and uniqueness.


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