Since W is Continuous Max W K w K Converges to 0 Partition

Much of analysis deals with the study ofR, the set of real numbers. It provides a rigourous foundation of concepts which we usually take for granted, e.g. continuity, differentiation, sequence convergence etc. One should have a mental picture of the set of rational numbersQ having "gaps" in its order structure, whileR "fills up" these gaps. We shall now describe this in a more rigourous manner.

First some basic definitions. SupposeX =R orQ andS is a non-empty subset ofX.

Definitions.

• Anupper bound of S is an element such that x ≥ y for all .If S has an upper bound, we say it is upper bounded.
• If is an upper bound of S, we call it themaximum of S, denoted max(S).
• Alower bound of S is an element such that x ≤ y for all . If S has a lower bound, we say it is lower bounded.
• If is a lower bound of S, we call it theminimum of S, denoted min(S).
• If S is both upper- and lower-bounded, we just say it isbounded.
• Let S be upper-bounded, and T the set of upper bounds of S. The minimum of T is called thesupremum of S, denoted sup(S).
• Let S be lower-bounded, and T the set of lower bounds of S. The maximum of T is called theinfimum of S, denoted inf(S).

Some basic properties include:

• The maximum (resp. minimum) is unique if it exists.
• Hence, the infimum (resp. supremum) is unique if it exists.
• If max(S) exists, then so does sup(S), and max(S) = sup(S).
• If min(S) exists, then so does inf(S), and min(S) = inf(S).
• A set may be upper-bounded without possessing a maximum, e.g. the set of real (or rational) numbers 0 <x < 1. Likewise, it may be upper-bounded without possessing a minimum.

A useful picture to visualise all these definitions is:

Example. The open interval (0, 1) is bounded but has no maximum or minimum. On the other hand, its infimum is 0 and supremum is 1. The closed interval [0, 1] has maximum (and hence supremum) 1.

Now we shall state the property which distinguishesR fromQ.

Completeness Property. Every upper-bounded subset S ofR has a supremum. We say thatR iscomplete.

ReplacingS with –S, we see that every lower-bounded subsetS ofR has an infimum. We're going to take this property for granted and proceed from here. By the way, to see that Q is not complete, consider the set of rational numbers r such thatr ² < 2. Clearly this is a bounded subset of Q, but it doesn't have a supremum in Q (and sinceR is complete, this set has a supremum inR, namely √2).

We end this section with a quick observation. By definitionL = sup(S) iffL is an upper bound ofS and there is no smaller upper boundL-ε (for ε>0). SinceL-ε is not an upper bound, there existsx inS,x >L-ε.

Thus, L = sup(S) iff (i) L is an upper bound, (ii) for each ε>0, there exists x in S, x > L-ε.

Definition of Convergence

A sequence in R is given by (a ₁, a ₂, a ₃, …), where each a_i  is in R. One can think of it as a function N → R, where N is the set of positive integers. Our focus here is to provide a rigourous foundation for the statement "sequence (a_n ) → L as n → ∞".

Definition (Convergence). Let (a_n) be a sequence. We say that (a_n) converges to a real number L (written a_n → L) if:

• for every positive real ε, there exists N, such that whenever n>N, we have |a_n – L| < ε.

We say a sequence isconvergent if it converges to some real L.

The definition takes some getting used to, so first we'll use this to prove that the limit, if exists, must be unique.

Example 0. If (a_n ) →K and (a_n ) →L, thenK=L.

Proof. IfK ≠L, take ε=|K–L|/2. By definition,

• we can findN such that whenn >N, we have |a_n –K| < ε;
• we can find M such that whenn >M, we have |a_n –L| < ε.

Then we can findn > max(M,N) such that

which is clearly absurd. ThusK=L. ♦

Example 1. Prove that if a _n= 1/n, then (a_n ) → 0.

Proof. For each ε>0, we need to establish anN such that whenn >N, we have |a_n  – 0| < ε. But |a_n | = 1/n, so this is easy: just setN = 1/ε. Now:

so by definition the sequence converges to 0. ♦

Example 2. Prove that ifa _n= 3n/(n+2), then (a_n ) → 3.

Proof. Let's first write . Now, for any ε>0, we setN = 6/ε. Then:

,

which shows that the sequence converges to 3. ♦

Example 3. Prove that the sequencea _n = (-1) ⁿ is not convergent.

Proof. We need to take the negation of the definition, i.e. what does it mean to say thata _n does not converge toL? There's a nice trick to take the negation of such statements:

• "It is not true that for everyx,P(x) holds." → "There existsx,P(x) doesn't hold."
• "It is not true that there existsx,P(x) holds." → "For anyx,P(x) doesn't hold."

Applying this rule-of-thumb recursively, we see that:

Tip. The sequence a_ndoes not converge to L if and only if:

• there exists an ε, such that for any N, we can find n>N satisfying |a_n-L| ≥ ε. [ Or equivalently, there exists an ε and infinitely many n such that |a_n-L| ≥ ε. ]

For our problem, suppose (a_n ) → L. Take ε = 1. IfL ≥ 0, then for any oddn, we have |a_n –L| = |-1-L| ≥ 1. And we're done since there're infinitely many oddn. On the other hand, ifL < 0, then pick the evenn's: we have infinitely many n for which |a_n –L| = |1-L| ≥ 1. ♦

Arithmetic Properties of Limits

For more complicated expressions ofa_n , it becomes a pain to consistently use the ε-N definition, so we need further properties to help us.

Theorem. Let be sequences converging to K, L respectively. Then:

1. converges to K+L;
2. a convergent sequence is bounded;
3. converges to KL;
4. if L ≠ 0, then converges to 1/L.

Proof. The proofs are conceptually easy but a bit tedious.

1. For addition: given ε>0, since ε/2>0, we can

• findM such that whenn >M, |a_n –K| < ε/2;
• find M' such that whenn > M', |b_n  –L| < ε/2.

Hence if we letN = max{M,M'}, then whenevern >N, we have:

2. Let's show is bounded. Since , we pick ε=1, and by definition of convergence, there existsN such that whenn >N, we have

Now there are only finitely many a_n  forn≤N, so we can letB be the maximum of these |a_n |. Then the set ofall |a_n | is bounded by max{B, |K|}.

3. To show , just use:

Now use the fact that |a_n | is bounded, say by B. Then: . Given any ε>0, pick M,M' such that:

• whenn>M, we have |a_n –K| < ε/(2|L|);
• whenn>M', we have |b_n –L| < ε/(2B).

Thus, lettingN = max{M,M'}, we see that whenevern>N, .

4. Finally, to show that , write:

.

Now pickM' such that whenever n>M', we have and so .

Suppose we're given ε>0. Since , there exists M such that whenevern>M, we get . Now letting N = max{M,M'}, whenevern>N, we get:

♦

Having proven these basics, some consequences are immediate.

Corollary. Let be sequences converging to K, L respectively. Then:

Proof.

For the first statement note that the constant sequence (c,c, … ) converges toc. Hence, the result follows by multiplying (a_n ) and (c). For the second statement, writea_n –b_n  =a_n  + (-1)b_n  and apply the first statement. For the third, writea_n /b_n  =a_n ·(1/b_n ). ♦

Now we can prove Example 2 using Example 1. Indeed, 3n/(n+2) = 3/(1+2/n). Since 1/n → 0 by Example 1, we have 2/n → 0. Thus 1+(2/n) → 1 and 3/(1+2/n) → 3, which is much easier.

Example 4. Prove thata _n= (n ² + 2n – 1)/(2n ² –n – 2) converges. Find its limit.

Answer. Divide the numerator and denominator by n ² to obtain . Use the fact that 1/n ² → 0 and we geta _n→ 1/2. ♦

Example 5. Find the limit of (2012 ⁿ + 2011 ⁿ )/(2012 ⁿ – 2011 ⁿ ) asn → ∞.

Answer.Divide the numerator and denominator by 2012 ⁿ  to obtain where α = 2011/2012. Since ,a _n→ 1. ♦

The final example would have been hellish to prove with the ε-N definition.

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Source: https://mathstrek.blog/2012/11/09/basic-analysis-sequence-convergence-1/

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