- #1

- 441

- 0

He says that if a = b then {a,b} will just be {a,a} = {a} and therefore will only be {{a}}.

What I don't understand is the the need for {a,b}, why not just define the Cartesian product to be {{a},{b}}. If a = b you get the same result.

- Thread starter Diffy
- Start date

- #1

- 441

- 0

He says that if a = b then {a,b} will just be {a,a} = {a} and therefore will only be {{a}}.

What I don't understand is the the need for {a,b}, why not just define the Cartesian product to be {{a},{b}}. If a = b you get the same result.

- #2

matt grime

Science Advisor

Homework Helper

- 9,395

- 3

The key thing is the notion of ordering. Sets do not come with any order on the elements. This is why you need this fiddle if you wish to define the cartesian product of sets purely in set theoretic terms.

- #3

- 441

- 0

So (a,b) = {{a},{a,b}} and (b, a) = {{b}, {b,a}}

Clearly (a,b) != (b,a)

Where as if you use my proposed definition you would have...

(a,b) = {{a},{b}} and (b,a) = {{b}, {a}}, but these are the same sets.

What is interesting is to me now is looking at the intersection and union of (a,b) and (b,a)

The intersection is {{a,b}}, the union is {{a},{b},{a,b}}. Cool!

On a side note Matt Grime, I don't know the correct quote in you signature, but you are clearly missing another essential tool of a mathematician... Coffee.

- #4

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

That's what I'm doing wrong; I don't drink coffee! :tongue:

- Last Post

- Replies
- 30

- Views
- 15K

Constructive Proofs
Munkre's Topology Ch 1 sec. 2, ex. #1:

- Last Post
- Math Proof Training and Practice

- Replies
- 1

- Views
- 243

- Last Post

- Replies
- 9

- Views
- 5K

- Last Post

- Replies
- 3

- Views
- 2K

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 9

- Views
- 5K

- Replies
- 3

- Views
- 927

- Last Post

- Replies
- 3

- Views
- 3K

- Last Post

- Replies
- 5

- Views
- 1K

- Replies
- 1

- Views
- 6K