Infinity. Suppose that $\{x_n\}$ is a sequence in $\R$. We say that $\lim_{n \rightarrow \infty} x_n = \infty$ if for any $M \in \R$, there exists $N \in \N$ such that for all $n \geq N$, $x_n \geq M$. The same convention is used analogously for $\lim_{n \rightarrow \infty} x_n = -\infty$.

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Subsequences

Given a sequence $\{x_n\}$ in a metric space $X$ and a strictly increasing sequence $\{n_j\}$ in $\N$, the sequence $\{x_{n_j}\}$ is called a subsequence of $\{x_n\}$.

Subsequence-Sequence Limit Characterization. A sequence $\{x_n\}$ converges to $x$ if and only if every subsequence $\{x_{n_j}\}$ converges to $x$.

Proof </aside>

Compact Metric Spaces contain Convergent Subsequences. Given a sequence $\{x_n\}$ in a compact metric space $X$, there exists a convergent subsequence of $\{x_n\}$.

Proof </aside>

Bolzano-Weierstrass Theorem. For a sequence $\{x_n\}$ with bounded range in $\R^k$, there exists a convergent subsequence of $\{x_n\}$.

Proof </aside>

Set of Subsequential Limits is Closed. For a sequence $\{x_n\}$ in any metric space $X$, the set $E$ of its subsequential limits is closed.

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lim-sup and lim-inf

Given a sequence $\{x_n\}$ in $\R$, let $E$ be the set of all limits (including limits in $\{-\infty, \infty\}$) of convergent subsequences of $\{x_n\}$. Then we define the lim-sup and lim-inf of $\{x_n\}$ as follows:

$$ \limsup _{n \rightarrow \infty} x_n := \sup E \\ \limsup _{n \rightarrow \infty} x_n := \inf E $$

If we accept the convention that if $E$ is unbounded above/below then its supremum/infimum is $+\infty$ or $-\infty$ respectively, this means that the lim-sup and lim-inf always exist. Moreover, there exists an alternative, equivalent definition for this:

$$ \limsup _{n\rightarrow \infty} x_n := \lim {n\rightarrow \infty } \sup{m \geq n} x_m \\ \liminf _{n \rightarrow \infty} x_n := \lim _{n\rightarrow \infty} \inf _{m \geq n}x_m $$

Limit Exists iff Upper and Lower Limits Agree. Given a sequence $\{x_n\}$ in $\R$, its limit exists in $\R \cup \{ -\infty, \infty\}$ iff $\limsup_{n\rightarrow\infty} x_n = \liminf_{n \rightarrow \infty}x_n$.

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Upper/Lower Limits Bound if Sequence Bound. If $\{s_n\}$ and $\{t_n\}$ are sequences such that for all $n \geq N$, for some fixed $N$, $s_n \leq t_n$, then their upper and lower limits are related in the same way.

$$ \exists N \in \N : \forall n \geq N , s_n \leq t_n \implies \begin{cases}\limsup {n\rightarrow \infty} s_n \leq \limsup{n\rightarrow \infty} t_n \\ \liminf {n\rightarrow \infty} s_n \leq \liminf{n\rightarrow \infty} t_n \end{cases} $$

As a corollary, the same holds true for their limits, if they exist.

$$ \exists N \in \N : \forall n \geq N , s_n \leq t_n \text{ and $s_n \rightarrow s$, $t_n \rightarrow t$ } \implies s \leq t $$

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$E$ contains $\sup E$ and there are Infinitely Many Terms less than $\beta > \sup E$. Given a sequence $\{a_n\}$ in $\R$, let $\alpha = \limsup_{n \rightarrow \infty} a_n$ .

$$ \alpha \in E \text{ and } \alpha < \beta \implies \exists N \in \N : \forall n \geq N, a_n < \beta $$

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