# Problem extending the notions of convergence.

Let $(x_n)_{n \in \mathbb{N}} \subset \mathbb{R}$ be a sequence. Let $x \in \mathbb{R}$. I know the following notions of

i) convergence

We write $\lim_{n \rightarrow \infty} x_n = x$ if $\forall \epsilon >0$ $\exists N \in \mathbb{N}$ such that $\forall n \geq N$ we have $|x_n-x| < \epsilon$.

ii) limsup (liminf analogously)

If $(x_n)_{n \in \mathbb{N}}$ is bounded, then write $\limsup_{k \rightarrow \infty} x_k := \lim_{k_0 \rightarrow \infty} \sup_{k \geq k_0} x_k$

Now under these notions I totally understand how one can construct for a bounded $(x_n)_{n \in \mathbb{N}}$ a convergent subsequence $(x_{n_k})_{k \in \mathbb{N}}$ such that $$\lim_{k \rightarrow \infty} x_{n_k}= \limsup_{n \rightarrow \infty} x_n$$

But now my questions are:

1. How does one extend the notion of convergence to the space $\overline{\mathbb{R}}= \mathbb{R} \cup \{-\infty, \infty\}$
2. What does it mean to say that $(x_n)_{n \in \mathbb{N}}$ converges to $-\infty$ or $\infty$?
3. How do I extend the definition of $\limsup x_n$ and $\liminf x_n$ if $(x_n)_{n \in \mathbb{N}}$ is unbounded?
4. How do I construct a subsequence of $(x_n)_{n \in \mathbb{N}}$ such that $(x_{n_k})_{n \in \mathbb{N}}$ converges to $\limsup x_n$ under this new notion if $\limsup x_n = \infty$

In my functional analysis class these notions were never extended. I am really thankful for any insight! Thanks in advance!

1. $\lim_{n\to\infty}x_n=+\infty$ means that$$(\forall M>0)(\exists p\in\mathbb{N})(\forall n\in\mathbb{N}):n\geqslant p\implies x_n>M$$ and $\lim_{n\to\infty}x_n=-\infty$ means that$$(\forall M>0)(\exists p\in\mathbb{N})(\forall n\in\mathbb{N}):n\geqslant p\implies x_n<-M.$$
3. If $(x_n)_{n\in\mathbb{N}}$ has no upper bound, then $\limsup_nx_n=+\infty$ and if $(x_n)_{n\in\mathbb{N}}$ has no lower bound, then $\liminf_nx_n=-\infty$.
4. $\limsup_nx_n=+\infty\iff(x_n)_{n\in\mathbb{N}}$ has no upper bound$\iff$ there is a subsequence $(x_{n_k})_{k\in\mathbb N}$ of $(x_n)_{n\in\mathbb N}$ such that $\lim_{k\to\infty}x_{n_k}=+\infty$.
• If I have an arbitrary sequence $(x_n)_{n \in \mathbb{N}}$ in $\mathbb{R}$. Then I know that there is a subsequences $(x_{n_k})$ such that it converges to $\liminf_{n \rightarrow \infty} x_n$, either if the sequence is bounded from below or it is not (the two senses of convergence now can happen). So if I somehow show that $\sup_{k \geq k_0} x_{n_k}$ is bounded from below by $C$ for every $k_0$ what is then $\limsup_{k \rightarrow \infty} x_{n_k}$? And in which sense? Is it equal to $\liminf_{n \rightarrow \infty} x_n$? And in what sense is now $\liminf_{n \rightarrow \infty} x_n$ understood? – vaoy Dec 1 '17 at 8:52
• @vaoy $\lim_{k_0\to\infty}\sup_{k\geqslant k_0}x_{n_k}=\limsup_kx_{n_k}$. – José Carlos Santos Dec 1 '17 at 9:21