2.5 Subsequences and cluster points (Page 4/4)

Analysis of functions of a single Page 4 / 4

Let ${a_{n}}$ be a bounded sequence of real numbers with

$lim sup a_{n} = L$ and $lim inf a_{n} = l .$ Prove that $L$ and $l$ satisfy the following properties.

For each $ϵ > 0,$ there exists an $N$ such that $a_{n} < L + ϵ$ for all $n \geq N .$ HINT: Use the fact that $lim sup a_{n} = L$ is the number $x$ of the lemma following Theorem 2.8, and that $x$ is the limit of a specific sequence ${x_{n}} .$
For each $ϵ > 0,$ and any natural number $k,$ there exists a natural number $j \geq k$ such that $a_{j} > L - ϵ .$ Same hint as for part (a).
For each $ϵ > 0,$ there exists an $N$ such that $a_{n} > l - ϵ$ for all $n \geq N .$
For each $ϵ > 0,$ and any natural number $k,$ there exists a natural number $j > k$ such that $a_{j} < l + ϵ .$
Suppose $L^{'}$ is a number that satisfies parts (a) and (b). Prove that $L^{'}$ is the limsup of ${a_{n}} .$ HINT: Use part (a) to show that $L^{'}$ is greater than or equal to every cluster point of ${a_{n}} .$ Then use part (b) to show that $L^{'}$ is less than or equal to some cluster point.
If $l^{'}$ is any number that satisfies parts (c) and (d), show that $l^{'}$ is the liminf of the sequence ${a_{n}} .$

Let ${a_{n}}$ and ${b_{n}}$ be two bounded sequences of real numbers, and write $L = lim sup a_{n}$ and $M = lim sup b_{n} .$ Prove that $lim sup (a_{n} + b_{n}) \leq lim sup a_{n} + lim sup b_{n} .$ HINT: Using part (a) of the preceding exercise, show that for every $ϵ > 0$ there exists a $N$ such that $a_{n} + b_{n} < L + M + ϵ$ for all $n \geq N,$ and conclude from this that every cluster point $y$ of the sequence ${a_{n} + b_{n}}$ is less than or equal to $L + M .$ This will finish the proof, since $lim sup (a_{n} + b_{n})$ is a cluster point of that sequence.
Again, let ${a_{n}}$ and ${b_{n}}$ be two bounded sequences of real numbers, and write $l = lim inf a_{n}$ and $m = lim inf b_{n} .$ Prove that $lim inf (a_{n} + b_{n}) \geq lim inf a_{n} + lim inf b_{n} .$ HINT: Use part (c) of the previous exercise.
Find examples of sequences ${a_{n}}$ and ${b_{n}}$ for which $lim sup a_{n} = lim sup b_{n} = 1,$ but $lim sup (a_{n} + b_{n}) = 0 .$

We introduce next another property that a sequence can possess. It looks very like the definition of a convergent sequence, but it differs in a crucial way, and thatis that this definition only concerns the elements of the sequence ${a_{n}}$ and not the limit $L .$

A sequence ${a_{n}}$ of real or complex numbers is a Cauchy sequence if for every $ϵ > 0,$ there existsa natural number $N$ such that if $n \geq N$ and $m \geq N$ then $| a_{n} - a_{m} | < ϵ .$

REMARK No doubt, this definition has something to do with limits. Any time there is a positive $ϵ$ and an $N,$ we must be near some kind of limit notion.The point of the definition of a Cauchy sequence is that there is no explicit mention of what the limit is. It isn't that theterms of the sequence are getting closer and closer to some number $L,$ it's that the terms of the sequence are getting closer and closer to each other. This subtle difference is worth some thought.

Prove that a Cauchy sequence is bounded. (Try to adjust the proof of [link] to work for this situation.)

The next theorem, like the Bolzano-Weierstrass Theorem, seems to be quite abstract, but it also turns out to be a very useful tool for proving theorems about continity, differentiability, etc.In the proof, the completeness of the set of real numbers will be crucial. This theorem is not true in ordered fields that are not complete.

Cauchy criterion

A sequence ${a_{n}}$ of real or complex numbers is convergent if and only if it is a Cauchy sequence.

If $l i m a_{n} = a$ then given $ϵ > 0,$ choose $N$ so that $| a_{k} - a | < ϵ / 2$ if $k \geq N .$ From the triangle inequality, and by adding and subtracting $a,$ we obtain that $| a_{n} - a_{m} | < ϵ$ if $n \geq N$ and $m \geq N .$ Hence, if ${a_{n}}$ is convergent, then ${a_{n}}$ is a Cauchy sequence.

Conversely, if ${a_{n}}$ is a cauchy sequence, then ${a_{n}}$ is bounded by the previous exercise. Now we use the fact that ${a_{n}}$ is a sequence of real or complex numbers.Let $x$ be a cluster point of ${a_{n}} .$ We know that one exists by the Bolzano-Weierstrass Theorem. Let us show that in fact this number $x$ not only is a cluster point but that it is in fact the limit of the sequence ${a_{n}} .$ Given $ϵ > 0,$ choose $N$ so that $| a_{n} - a_{m} | < ϵ / 2$ whenever both $n$ and $m \geq N .$ Let ${a_{n_{k}}}$ be a subsequence of ${a_{n}}$ that converges to $x .$ Because ${n_{k}}$ is strictly increasing, we may choose a $k$ so that $n_{k} > N$ and also so that $| a_{n_{k}} - x | < ϵ / 2 .$ Then, if $n \geq N,$ then both $n$ and this particular $n_{k}$ are larger than or equal to $N .$ Therefore, $| a_{n} - x | \leq | a_{n} - a_{n_{k}} | + | a_{n_{k}} - x | < ϵ .$ this completes the proof that $x = lim a_{n} .$

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Source: OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1

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