I have some confusion on the definition of upper limit of a sequence. Usually we see this definition:

Let $(x_n)$ be a bounded sequence and for each natural number $n$, let $$\overline{x_n}=\sup \left \{ x_k:k\ge n \right \}$$ Then $\overline{x}=\lim \overline{x_n}$ is the upper limit of the sequence $(x_n)$.

However, in Rudin's Priciple of Mathemetical Analysis, he wrote:

Let $(s_n)$ is the sequence of real numbers. Ler $E$ be the set of numbers $x$ (in the extended real system) such that $s_{n_k}\rightarrow x$ for some subsequence $(s_{n_k})$. Define $$s^*=\sup E$$ then $s^*$ is called the upper limit of the sequence $(s_n)$.

I'm confused with the two definitions. I cannot find any connection between them. When doing exercises, I usually use the first definition, but how those two definitions are the same?

Thanks a lot.


These two definitions are actually equivalent:

Let $(x_n)$ be a sequence of real numbers. Define $$ a = \lim_{n \to \infty} \sup \{ x_k \mid k \geq n \} =: \lim_{n \to \infty} a_n \,, $$ and $$ b = \sup E\,, $$ where $E$ is the set of numbers $x \in \Bbb{R} \cup \{+ \infty, - \infty\}$ such that $x_{n_k} \to x$ for some subsequence $(x_{n_k})$. The numbers $a$ and $b$ are well defined numbers in $\Bbb{R} \cup \{+ \infty, - \infty\}$ since $a$ is a limit of a monotone sequence and $b$ is a supremum of a non-empty set (every sequence of $\Bbb{R} \cup \{+ \infty, - \infty\}$ has converging subsequence because $\Bbb{R} \cup \{+ \infty, - \infty\}$ is compact).

We want to show that $a = b$.

First, assume that $a < b$ holds. Now $\lim a_n < \sup E$. Because $(a_n)$ is a decreasing sequence, there exists a $n_0 \in \Bbb{N}$ s.t. $$ n \geq n_0 \implies a_n < \sup E \iff \sup\{x_k \mid k \geq n \} < \sup E $$ By definition of supremum we know that for every $\epsilon > 0$ there exists a subsequence $(x_{n_k})$ s.t. $x_{n_k} \to x$ and $\sup E - \varepsilon < x \leq \sup E$. So we can choose $\varepsilon = \sup E - a_{n_0} > 0$ and now we have a subsequence $(x_{n_k})$ converging to $x$ and $a_{n_0} < x \leq \sup E$. But it must also hold that if $n \geq n_0$, then $x_{n_k} \leq a_{n_0} < x$ so $\lim x_{n_k} \leq a_{n_0} < x = \lim x_{n_k}$, which is a contradiction.

Now assume that $a > b$. Then $\lim a_n > \sup E$. Now $a_n > \sup E$ for all $n$ because $(a_n)$ is decreasing. So for all $\varepsilon > 0$ and $n$ we can find $k_n \geq n$ s.t. $a_n - \varepsilon < x_{k_n} \leq a_n$. Now choose $\varepsilon = (a_n - \sup E)/2$ and we have $$ a_n - \varepsilon = \frac{a_n + \sup E}{2} < x_{k_n} < a_n\,. $$ Now, again because $\Bbb{R} \cup \{+ \infty, - \infty\}$ is compact, the sequence $(x_{n_k})$ has a converging subsequence which converges to $x$. By the previois inequalities, it must hold that $$ \sup E=\frac{ \sup E + \sup E}{2} < \frac{a_n + \sup E}{2} \leq x < a_n $$ $\sup E < x$ is a contradiction.

So $a = b$.


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