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<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)}.

— 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 […]

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