# Questions on Real Analysis

1. Dec 20, 2009

### Shackleford

I recently bought Real Analysis by Haaser and Sullivan. Is this a good introductory real analysis book? I really bought it for fun. I'm not going to put a formal real or complex analysis course in my math minor.

Well, I'm on page two at the ordered pair proof.

Why is an ordered pair defined as {{a}, {a, b}} ?

(a, b) = (c, d) if a = c and b = d.

Either {a} = {c} and {a, b} = {c, d}

or

{a} = {c, d} and {c} = {a, b}

I understand the second one. If the two sets are equal, then all of the elements in both sets are the same. So, I guess it's irrespective of the number of elements in each set as long as they're all the same.

For the first one, a = c, and b = c or d. Obviously, if b = d the proof is satisfied. However, why does b equal c or d?

2. Dec 20, 2009

### Hurkyl

Staff Emeritus
Because it's a simple set-theoretic model of the notion of ordered pair.

All we care about ordered pairs is:
• If P is a pair, then we have an object First(P) and an object Second(P)
• For any two objects x and y, there is a pair with First(P)=x and Second(P)=y
• Two pairs are equal iff their first objects are equal and their second objects are equal

It doesn't matter what ordered pairs "really are" -- these properties are the only way in which we will ever use them, so nothing else matters.

So, someone came up with this nifty model of ordered pairs so that we only have to study "set theory" rather than having to develop a separate and independent "ordered pair theory".

Anyways, the key properties about sets that you don't seem to be using is that if S and T are equal sets, then
• If s is in S, then s is in T
• If t is in T, then t is in S

Also, {a,b} is, by definition, the set with the property that x is in {a,b} if and only if x is a or x is b.

3. Dec 20, 2009

### Shackleford

As far as I can remember, this is my first exposure to set theory. The book certainly assumes knowledge of the basic properties of sets. Do you know of a good website that covers that?

So, for the last question about {a, b} = {c, d}, you can assume a = c or d and b = c or d and vice versa since the sets are equal. You're basically just playing with all the possible combinations?

4. Dec 20, 2009

### JSuarez

I don't know that particular text, so I can't comment on it, but Analysis textbooks are pretty much standard these days. Just one thing: if you are taking a math minor, I would definitely advise you to take, at least, a Real Analysis course; Complex Analysis would probably be more useful, but it assumes that you know a number of concepts from the real counterpart.

Regarding ordered pairs, there is reason to define them like that: for the mathematician, it's important to prove that mathematical objects can be reduced to the ones that are presently seen as foundational, which now are sets.
That definition is due to Kuratovski and it dates back to the beginning of the XXth century, and is the simplest one in terms of sets.
Of course, the persons who only uses ordered pairs doesn't have to bother how they are defined, but only with their properties; in this case the defining property is equality:

(a,b) = (c,d) if and only if a = c and b = d (notice that you forgot one of the implications).

The problem is now to prove that the proposed definition does have this property; for this, you should also be aware that set equality is defined by extension: two sets are equal iff they have the same elements; this implies, for example, that {a} = {a,a}.

Then, to prove that {{a},{a,b}} = {{c},{c,d}} implies that a = c and b = d (the other implication is easier), you have to start from:

$$\left\{a\right\} \in \left\{\left\{c\right\},\left\{c,d\right\}\right\} \Rightarrow \left\{a\right\} = \left\{c\right\} or \left\{a\right\} = \left\{c,d\right\}$$

$$\left\{a,b\right\} \in \left\{\left\{c\right\},\left\{c,d\right\}\right\} \Rightarrow \left\{a,b\right\} = \left\{c\right\} or \left\{a,b\right\} = \left\{c,d\right\}$$

These indeed imply that a = c and b = d; regarding your specific question, note that, from
{a} = {c} and {a,b} = {c,d} you may infer that a = c (from the first), but not b = d; you just know that b is an element of {c,d}, so it may be equal to c OR d (if it's equal to c, then a = c = b and the proof also follows).

5. Dec 23, 2009

### Shackleford

Thanks, guys.

The definition of the union and intersection of a collection of sets is confusing me a bit.

"C" = collection of sets

Union = { x is an element of X : x is an element of A for some A is an element of C }
Intersection = { x is an element of X : x is an element of A for all A is an element of C }

I'm not exactly sure what this is saying. Now, I realize these are very elementary questions and I'm thinking maybe I should work my way up to this book. I only just finished Differential Equations this semester. What do you guys think? Should I trudge my way through this or start with some other prerequisite maths starting after DE? For my physics major, I only have to take Vector Analysis and Introduction to Partial Differential Equations then some 4000-level MATH I choose.

This is the book, btw.

https://www.amazon.com/Real-Analysi...5097/ref=sr_1_1?ie=UTF8&qid=1261623859&sr=8-1

Last edited by a moderator: Apr 24, 2017
6. Jan 4, 2010

### Shackleford

Well, I picked up this book at half price for \$7.98, Intermediate Mathematical Analysis, by Anthony Labarre, Jr. It appears as though it will give me a bit of necessary background knowledge to help better understand the Real Analysis book I have.

I also bought Vector Analysis by Homer E. Newell, Jr. for my Vector Analysis course in the spring. I'm not so sure the professor uses a textbook for that class, so that's why I bought it.