Much of analysis deals with the study of **R**, 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 numbers **Q** having “gaps” in its order structure, while **R** “fills up” these gaps. We shall now describe this in a more rigourous manner.

First some basic definitions. Suppose *X* = **R** or **Q** and *S* is a non-empty subset of *X*.

Definitions.

- An
upper boundof S is an element such that x ≥ y for all .If S has an upper bound, we say it isupper bounded.- If is an upper bound of S, we call it the
maximumof S, denoted max(S).- A
lower boundof S is an element such that x ≤ y for all . If S has a lower bound, we say it islower bounded.- If is a lower bound of S, we call it the
minimumof S, denoted min(S).- If S is both upper- and lower-bounded, we just say it is
bounded.- Let S be upper-bounded, and T the set of upper bounds of S. The minimum of T is called the
supremumof S, denoted sup(S).- Let S be lower-bounded, and T the set of lower bounds of S. The maximum of T is called the
infimumof 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 distinguishes **R** from **Q**.

Completeness Property. Every upper-bounded subset S ofRhas a supremum. We say thatRiscomplete.

Replacing *S* with –*S*, we see that every lower-bounded subset *S* of **R** 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 that *r*^{2} < 2. Clearly this is a bounded subset of **Q**, but it doesn’t have a supremum in **Q** (and since **R** is complete, this set has a supremum in **R**, namely √2).

We end this section with a quick observation. By definition *L* = sup(*S*) iff *L* is an upper bound of *S* and there is no smaller upper bound *L*-ε (for ε>0). Since *L*-ε is not an upper bound, there exists *x* in *S*, *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})convergesto 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 is

convergentif 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*, then

*K*=

*L*.

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

- we can find
*N*such that when*n*>*N*, we have |*a*–_{n}*K*| < ε; - we can find
*M*such that when*n*>*M*, we have |*a*–_{n}*L*| < ε.

Then we can find *n* > max(*M*, *N*) such that

which is clearly absurd. Thus *K*=*L*. ♦

**Example 1**. Prove that if *a*_{n }= 1/*n*, then (*a _{n}*) → 0.

**Proof**. For each ε>0, we need to establish an *N* such that when *n* > *N*, we have |*a _{n}* – 0| < ε. But |

*a*| = 1/

_{n}*n*, so this is easy: just set

*N*= 1/ε. Now:

so by definition the sequence converges to 0. ♦

**Example 2**. Prove that if *a*_{n }= 3*n*/(*n*+2), then (*a _{n}*) → 3.

**Proof**. Let’s first write . Now, for any ε>0, we set *N* = 6/ε. Then:

,

which shows that the sequence converges to 3. ♦

**Example 3**. Prove that the sequence *a*_{n}= (-1)^{n} is not convergent.

**Proof**. We need to take the negation of the definition, i.e. what does it mean to say that *a*_{n }does not converge to *L*? There’s a nice trick to take the negation of such statements:

- “It is not true that for every
*x*,*P*(*x*) holds.” → “There exists*x*,*P*(*x*) doesn’t hold.” - “It is not true that there exists
*x*,*P*(*x*) holds.” → “For any*x*,*P*(*x*) doesn’t hold.”

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

Tip. The sequence a_{n }does 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. If

*L*≥ 0, then for any odd

*n*, we have |

*a*–

_{n}*L*| = |-1-

*L*| ≥ 1. And we’re done since there’re infinitely many odd

*n*. On the other hand, if

*L*< 0, then pick the even

*n*‘s: we have infinitely many

*n*for which |

*a*–

_{n}*L*| = |1-

*L*| ≥ 1. ♦

## Arithmetic Properties of Limits

For more complicated expressions of *a _{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:

- converges to K+L;
- a convergent sequence is bounded;
- converges to KL;
- 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

- find
*M*such that when*n*>*M*, |*a*–_{n}*K*| < ε/2; - find
*M’*such that when*n*>*M’*, |*b*–_{n}*L*| < ε/2.

Hence if we let *N* = max{*M*, *M’*}, then whenever *n* > *N*, we have:

2. Let’s show is bounded. Since , we pick ε=1, and by definition of convergence, there exists *N* such that when *n* > *N*, we have

Now there are only finitely many *a _{n}* for

*n*≤

*N*, so we can let

*B*be the maximum of these |

*a*|. Then the set of

_{n}*all*|

*a*| is bounded by max{

_{n}*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:

- when
*n*>*M*, we have |*a*–_{n}*K*| < ε/(2|*L*|); - when
*n*>*M’*, we have |*b*–_{n}*L*| < ε/(2*B*).

Thus, letting *N* = max{*M*, *M’*}, we see that whenever *n*>*N*, .

4. Finally, to show that , write:

.

Now pick *M’* such that whenever *n*>*M’*, we have and so .

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

♦

Having proven these basics, some consequences are immediate.

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

- for any real c, converges to cK;
- converges to K-L;
- if L ≠ 0, then converges to K/L.

**Proof**.

For the first statement note that the constant sequence (*c*, *c*, … ) converges to *c*. Hence, the result follows by multiplying (*a _{n}*) and (

*c*). For the second statement, write

*a*–

_{n}*b*=

_{n}*a*+ (-1)

_{n}*b*and apply the first statement. For the third, write

_{n}*a*/

_{n}*b*=

_{n}*a*·(1/

_{n}*b*). ♦

_{n}Now we can prove Example 2 using Example 1. Indeed, 3*n*/(*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 that *a*_{n }= (*n*^{2} + 2*n* – 1)/(2*n*^{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 get *a*_{n }→ 1/2. ♦

**Example 5**. Find the limit of (2012^{n} + 2011^{n})/(2012^{n} – 2011^{n}) as *n* → ∞.

**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.