Real Analysis – Sequences II

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 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:

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 ${ \forall \delta > 0 \, \exists k \in \mathbb{N}: \, n \geq k \Rightarrow u_n \in V(a,\delta)}$.

— 5.2. Limits and Inequalities —

 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 ${ \lim \dfrac{\sin n}{n}=0}$.

3 Responses to “Real Analysis – Sequences II”

1. […] 3, 2009 After introducing sequences and gaining some knowledge of some of their properties (I,II ,III , and IV) we are ready to embark on the study of real analysis while using concepts that are […]

2. […] Theorem 17 in post Real Analysis – Sequences II there exists a sequence of points in such that […]

3. […] by the Squeezed Sequence Theorem […]