Dictionary order and least upper bound property

[Samuelb88]

Homework Statement

Does [0,1] \times [0,1] in the dictionary order have the least upper bound property?

Homework Equations

Dictionary Order. (on \mathbb{R}^2) Let x , y \in \mathbb{R}^2 such that x=(x_1 , x_2) and y = (y_1 , y_2). We say that x < y if x_1 < y_1, or if x_1 = y_1 and x_2 < y_2.

Def'n. An ordered set A is said to have the least upper bound property if every nonempty subset A_0 \subseteq A that is bounded has a least upper bound.

Assume that the real line has the least upper bound property.

The Attempt at a Solution

I'm not sure if I am proving this correctly. Here's my proof.

I want to show that every subset A_0 \subseteq [0,1] \times [0,1] that is nonempty and bounded has the lub property. Suppose that \mathbb{R} has the lub property. Let A_0 be an nonempty subset of [0,1] \times [0,1]. Since [0,1] \times [0,1] is bounded, it follows that every subset of [0,1] \times [0,1] is bounded. We will consider two forms of A_0, that is, when either A_0 = [i,j] \times [k, \ell], or A_0 = (i,j) \times (k, \ell), where 0 \leq i, j, k, \ell \leq 1.

If A_0 = [i,j] \times [k, \ell], then \forall x \in A_0, we can always find a least upper bound, say y by letting y=x. So that case is settled.

Instead, suppose that A_0 = (i,j) \times (k, \ell). Then \forall x \in A_0, we can still always find a least upper bound which we will again call y such that y = (y_1 , y_2) by letting y_1 = j and y_2 = k.

In a similar manner, we can show that subsets that have both closed and open ends (e.g. (i,j] \times (k, \ell]) always have a least upper bound.

Therefore I have shown that every subset of [0,1] \times [0,1] that is nonempty and bounded has the lub property and therefore the set [0,1] \times [0,1] has the lub property.

How does this look?

 

[tiny-tim]

Hi Samuelb88!

But you're only proving it for open and closed squares …

what about open and closed triangles, and more complicated regions?

 

[Samuelb88]

Okay, given A_0 \exists i, j, k, \ell \in \mathbb{N} such that A_0 \subseteq [i,j] \times [k,\ell]. In words, I'm claiming that there exists a box with dimensions |j-i| and |\ell - k| such that the region of A_0 lies entirely inside the box. Would this help? I'm not allowed to use the concept of a border of a region so I'm not sure how to explain how we can fit the box about either an open or closed region so that the border of the box coincides with the "border" of the region (By "border" of the region, I mean the line that encloses it, whether it be dashed (open region) or solid (closed region)).

What if I let b \in [i,j] \times [k,\ell] such that b = (k,\ell). Then b is an upperbound in the dictionary order for the region of A_0. Now look at the set of all such upper bounds of the region A_0, that is, \{ b \in [i,j] \times [k,\ell] : \forall x \in A_0 \, \, x \leq b\}. Denote this set of upper bounds by U_{A_0}. Now in U_{A_0} there will be a an element, say b_0, such that b_0 \leq b \forall b \in [i,j] \times [k,\ell] in the dictionary order. Basically I want to let b_0 = \min(U_{A_0}), but I am not sure if I'm using the min function correctly. At any rate, this b_0 is the lub of the set A_0. Therefore (since A_0 was arbitrary) each subset of [0,1] \times [0,1] has the lub property in the dictionary order and therefore by our proposition stated in my original post, [0,1] \times [0,1] has the lub property in the dictionary order.

 

[micromass]

Okay, given A_0 \exists i, j, k, \ell \in \mathbb{N} such that A_0 \subseteq [i,j] \times [k,\ell]. In words, I'm claiming that there exists a box with dimensions |j-i| and |\ell - k| such that the region of A_0 lies entirely inside the box. Would this help? I'm not allowed to use the concept of a border of a region so I'm not sure how to explain how we can fit the box about either an open or closed region so that the border of the box coincides with the "border" of the region (By "border" of the region, I mean the line that encloses it, whether it be dashed (open region) or solid (closed region)).

Well, of course such a box exists, take

[i,j]\times[k,l]=[0,1]\times [0,1]

but that won't help us.

Given a set A₀. Can you prove that

\{x~\vert~\exists y:~(x,y)\in A_0\}

has an upper bound? I.e., when you project the set onto the x-axis, does it have an upper bound then?

 

[Samuelb88]

micromass,

If I am understanding your question correctly, then x=1 is one such upper bound for the set \{ x : \exists y : (x,y) \in A_0 \}, which I will denote as U_{A_0}. So then can I look at the set of all upper bounds of U_{A_0} and take the \min(U_{A_0}) to show that it has a least upper bound?

 

[micromass]

micromass,

If I am understanding your question correctly, then x=1 is one such upper bound for the set \{ x : \exists y : (x,y) \in A_0 \}, which I will denote as U_{A_0}. So then can I look at the set of all upper bounds of U_{A_0} and take the \min(U_{A_0}) to show that it has a least upper bound?

Indeed, so U_{A_0} has a least upper bound x_0. Can you use this to find an upper bound of A_0?? Maybe look at \{y~\vert~(x_0,y)\in A_0\}.

 

[Samuelb88]

Okay, letting x_0 be the least upper bound of U_{A_0}. I can play the same game with the set V = \{ y : (x_0 , y) \in A_0 \}. I know that one such upper bound of the set V is y=1. Thus looking at the set of all upper bounds of V, I will again take the minimum of that set and have found an upper bound, say y_0. Thus I have found an upper bound of A_0, that is (x_0 , y_0).

 

[micromass]

Okay, letting x_0 be the least upper bound of U_{A_0}. I can play the same game with the set V = \{ y : (x_0 , y) \in A_0 \}. I know that one such upper bound of the set V is y=1. Thus looking at the set of all upper bounds of V, I will again take the minimum of that set and have found an upper bound, say y_0. Thus I have found an upper bound of A_0, that is (x_0 , y_0).

And can you show that (x_0,y_0) is the least upper bound?

 

[Samuelb88]

Well it seems to me from our construction of x_0 and y_0 that (x_0,y_0) is the least upper bound of A_0 is it not?

 

[micromass]

Well it seems to me from our construction of x_0 and y_0 that (x_0,y_0) is the least upper bound of A_0 is it not?

Well, if it's obvious to you, then the proof is finished!

 

[Samuelb88]

Great! Thanks, micromass! Always appreciated. :)