# How is it defined that $a_1 \times b_1 < a_2 \times b_1$ and that $a_1 \times b_3 < a_3 \times b_1$?

I am reading Topology by James Munkres and he defines the dictionary order relation as:

Definition Suppose that $$A$$ and $$B$$ are both sets with order relations $$<_A$$ and $$<_B$$ respectively. Define an order $$<$$ on $$A\times B$$ by defining

$$a_1 \times b_1 < a_2 \times b_2$$

if $$a_1<_A a_2$$, or $$a_1=a_2$$ and $$b_1<_B b_2$$. It is called the dictionary order relation on $$A\times B$$.

I think that I intuitively understand the relation as working as indexing words in the dictionary. The problem is that I do not understand how is it that the order relation of the following cases is defined in the previous definition:

1. How is it in the definition that $$a_1 \times b_1 < a_2 \times b_1$$ ?
2. How is it in the definition that $$a_1 \times b_3 < a_3 \times b_1$$ ?

I kind of intuitively feel that it could be deduced from the definition but I do not see how. I want to understand it so I make sure I do understand the concept and the definition.

Thanks

• Points in $A\times B$ are compared by their first elements, and the second elements are compared if there is a "tie" with the first elements. If the order can be determined using only the first elements, the second elements aren't used. For example, if $A=B=\mathbb R$, then $(1,r)<(2,s)$ no matter what $r$ and $s$ are because $1<_A 2$. – MPW May 18 at 13:23

If you look at the definition of dictionary order relation, you see that you can say in words that $$(a_1, b_1) < (a_2, b_2)$$ if and only if

• The first element of the pair $$(a_1, b_1)$$ is less than the first element of the pair $$(a_2, b_2)$$ with respect to $$<_A$$, or
• The first two elements of the pairs $$(a_1, b_1)$$ and $$(a_2, b_2)$$ are equal and the second element of the pair $$(a_1, b_1)$$ is less than the second element of the pair $$(a_2, b_2)$$ with respect to $$<_B$$.

With this, we can see that the defintion for $$(a_1, b_1) < (a_2, b_1)$$ is just:

• $$a_1 <_A a_2$$, or
• $$a_1 = a_2$$ and $$b_1 <_B b_1$$.

Note that in this case we don't have that $$b_1 <_B b_1$$ (since trivially $$b_1 = b_1$$), so if you know that $$(a_1, b_1) < (a_2, b_1)$$, then the definition of the dictionary order relation implies that $$a_1 <_A a_2$$.

For $$(a_1, b_3) < (a_3, b_1)$$ the definition is that:

• $$a_1 <_A a_3$$, or
• $$a_1 = a_3$$ and $$b_3 <_B b_1$$.

In this case, if you know that $$(a_1, b_3) < (a_3, b_1)$$, then the defintion of the dictionary order relation implies that one of the above has to hold; if no more information is given about $$a_1,a_3,b_1$$ and $$b_3$$ then no more information can be deduced.

• Hhmm I see, that makes sense. My problem was the comma in the explanation of the iff. I thought that it was (𝑎1 <𝐴 𝑎2 and 𝑏1 <𝐵 𝑏2) or (𝑎1 = 𝑎2 and 𝑏1 <𝐵 𝑏2) thanks! – César D. Vázquez May 20 at 11:16
• @CésarD.Vázquez ¡De nada! – Rick May 20 at 11:19