Willard, Ordered Pair problem

[malicx]

Homework Statement

Show that, if (x₁, x₂) is defined to be {{x₁. {x₁, y₂}}, then (x₁, x₂) = (y₁, y₂) iff x₁ = x₂ and y₁ = y₂.

This is from Willard's General Topology, problem 1C.

I think Willard is trying to develop the set theoretic definition of the ordered pair, but this doesn't seem correct to me... In particular, it seems like we should be showing x₁ = y₁, etc. Is this is a gigantic typo or are we trying to show something completely different than what I'm assuming? In fact, even the definition seems to be wrong looking at http://planetmath.org/encyclopedia/OrderedPair.html . Note that I copied this exactly from the book.

 

[jbunniii]

Homework Statement

Show that, if (x₁, x₂) is defined to be {{x₁. {x₁, y₂}}, then (x₁, x₂) = (y₁, y₂) iff x₁ = x₂ and y₁ = y₂.

Do you mean

$$(x_1, x_2)$$ is defined to be $$\{x_1, \{x_1, x_2\}\}$$?

If so, that definition seems fine. One direction is trivial: if $x_1 = x_2$ and $y_1 = y_2$ then clearly $(x_1, x_2) = (y_1, y_2)$.

Conversely, if $(x_1, x_2) = (y_1, y_2)$ then by definition

$$\{x_1, \{x_1, x_2\}\} = \{y_1, \{y_1, y_2\}\}$$

Each set has two elements, and the elements are of different types: a singleton and a set of two elements. What does equality imply?

 

[malicx]

Do you mean

$$(x_1, x_2)$$ is defined to be $$\{x_1, \{x_1, x_2\}\}$$?

If so, that definition seems fine. One direction is trivial: if $x_1 = x_2$ and $y_1 = y_2$ then clearly $(x_1, x_2) = (y_1, y_2)$.

Conversely, if $(x_1, x_2) = (y_1, y_2)$ then by definition

$$\{x_1, \{x_1, x_2\}\} = \{y_1, \{y_1, y_2\}\}$$

Each set has two elements, and the elements are of different types: a singleton and a set. What does equality imply?

I know how to prove it given that definition. I copied it exactly from the book. Take a look here, it's on page 13 http://books.google.com/books?id=-o...&resnum=3&ved=0CCsQ6AEwAg#v=onepage&q&f=false.

 

[jbunniii]

I know how to prove it given that definition. I copied it exactly from the book. Take a look here, it's on page 13 http://books.google.com/books?id=-o...&resnum=3&ved=0CCsQ6AEwAg#v=onepage&q&f=false. It's got to be a typo. Thanks.

Yes, I agree it's a typo. The definition makes no sense with $y_2$ instead of $x_2$.

 

[malicx]

Yes, I agree it's a typo. The definition makes no sense with $y_2$ instead of $x_2$.

Haha that's quite frustrating. I've found other typos in the book as well, and people seem to be praising it. I guess I'll just be extra careful