After having introduced the notion of neighborhood of a real number in the previous post we are going to introduce a further notion that will permit us to unify a few results.

Definition 16 The set formed by $ { \{ -\infty \} \cup \mathbb{R} \cup \{\infty \}}$ will be the set of affinely extended real numbers and denote it by $ { \overline{\mathbb{R}}}$. The elements of this new set have the following properties: $ { x\in\mathbb{R}\Rightarrow -\infty<x<\infty}$ and $ { -\infty<\infty}$ $ { x+\infty=\infty}$ if $ { x\neq -\infty}$ $ { x-\infty=-\infty}$ if $ { x\neq \infty}$ $ { x\cdot(\pm\infty)=\pm\infty}$ if $ { x>0}$ $ { x\cdot(\pm\infty)=\mp\infty}$ if $ { x<0}$ $ { \dfrac{x}{\pm\infty}=0}$ if $ { x\neq\pm\infty}$ $ { \displaystyle\left|\frac{x}{0}\right|=\infty}$ if $ { x\neq 0}$

After the introduction of this new set and the new elements we can define two new neighborhoods:

Definition 17 $ \displaystyle V(+\infty,\delta)=\left\rbrack\dfrac{1}{\delta},+\infty\right\rbrack \ \ \ \ \ (13)$ $ \displaystyle V(-\infty,\delta)=\left\lbrack -\infty,-\dfrac{1}{\delta}\right\lbrack \ \ \ \ \ (14)$

It is very important that the reader can understand the reasons of these two definitions. I know that the first time I looked at them I was puzzled.

The notion of neighborhood of a real number is straightforward. Essentially, it is the open interval centered at a given number.

If one wants to extend the notion of neighborhood to the new elements $ { +\infty}$ and $ { -\infty}$ one has to realize that one has to abandon the hope of getting a centered interval. This can't be done because those two new elements are the edges of $ { \overline{\mathbb{R}}}$.

One thing that one can, and should keep, though, is the fact that the bigger the $ { \delta}$ the bigger the interval one gets. But with $ { 1/\delta}$ one gets smaller and smaller ratios as $ { \delta}$ gets bigger!!! And that's just want we want! If $ { 1/\delta}$ gets smaller this means that the left edge of the neighborhood (I'm assuming we are talking about the $ { +\infty}$ case) is moving even more to the left, thus making the interval bigger!

The neighborhood for $ { -\infty}$ has the same reasoning behind it and I hope that now any doubts that you might have had regarding the definitions of the neighborhoods of $ { +\infty}$ and $ { -\infty}$ are gone.

Let us now consider a sequence, $ { u_n}$, that is bounded below but not bounded above. That is to say that in $ { \overline{\mathbb{R}}}$ we have that $ { u_n \rightarrow +\infty}$. This is equivalent to the following:

$ { \begin{array}{rcl} \forall \delta > 0 \,\exists k \in \mathbb{N}: \, n\geq k \Rightarrow u_n > \dfrac{1}{\delta} &\Leftrightarrow& u_n \in \left\rbrack \dfrac{1}{\delta}, +\infty\right\rbrack \\ &\Leftrightarrow& u_n \in V(+\infty,\delta) \end{array}}$

One can do the analogous derivation for $ { u_n \rightarrow -\infty}$. Hence one can write with full generality:

Definition 18 Given $ { a \in \overline{\mathbb{R}}}$ it is $ { \lim u_n=a}$ iff $ { \forall \delta > 0 \, \exists k \in \mathbb{N}: \, n \geq k \Rightarrow u_n \in V(a,\delta)}$.

Theorem 14 Given $ { u_n}$ and $ { v_n}$ let us suppose that there exists an order $ { m}$ such as $ { u_n \leq v_n \quad \forall n \geq m}$. If $ { \lim u_n}$ and $ { \lim v_n}$ exist, it follows $ { \lim u_n \leq \lim v_n}$ Proof: Omitted. $ \Box$

After this theorem we may ask ourselves is $ { u_n < v_n \Rightarrow \lim u_n < \lim v_n}$ also a theorem? Well, it isn't. To prove so a single counterexample suffices.
For instance $ { \dfrac{1}{n+1}<\dfrac{1}{n} \quad \forall n\geq 1}$ and nevertheless $ { \lim \dfrac{1}{n+1}= \lim \dfrac{1}{n} = 0}$.

Corollary 15 Let $ { u_n}$ be a sequence and $ { a \in \mathbb{R}}$. Let us also suppose that from a certain order we have $ { u_n \leq a}$ ($ { u_n \geq a}$), then if $ { \lim u_n}$ exists we have $ { \lim u_n \leq a}$ ( $ { \lim u_n \geq a}$). Proof: Take $ { v_n=a\quad \forall n}$ in theorem 14. $ \Box$

Theorem 16 Given $ { u_n}$ and $ { v_n}$ such as $ { u_n \leq v_n \quad \forall n > m}$ for some order $ { m}$. Then: $ { u_n \rightarrow +\infty \Rightarrow v_n \rightarrow +\infty}$ $ { v_n \rightarrow - \infty \Rightarrow u_n \rightarrow -\infty}$ Proof: Omitted. $ \Box$

Theorem 17 (Squeezed sequence theorem) Given $ { u_n}$, $ { v_n}$, and $ { w_n}$ such as, for some order $ { m}$, $ { v_n \leq u_n \leq w_n}$. Then, if $ { \lim v_n = \lim w_n}$, $ { u_n}$ has a limit and it is $ { \lim v_n = \lim u_n = \lim w_n}$. Proof: Omitted. $ \Box$

As an application of theorem 17 let us look at the following example: $ { u_n=\dfrac{\sin n}{n}}$ and we wish to compute $ { \lim u_n}$.

Well, $ { -1\leq \sin n \leq 1 \Rightarrow -\dfrac{1}{n} \leq \dfrac{\sin n}{n} \leq \dfrac{1}{n}}$.
We know that $ { \lim\left( -\dfrac{1}{n} \right)= \lim \dfrac{1}{n} = 0}$, hence $ { \lim \dfrac{\sin n}{n}=0}$.