Question

A = set of all English words. R = {(x,y)∈AxA | the word y occurs at least as late in alphabetical order as the word x}. What does this even mean???

Answer 1

You should think of R as a type of ordering relation, like \(<,\leq,>\geq \)

Answer 2

In a way it is basically \(x \geq y\), where every word is sorted in alphabetical order, and \(x\) and \(y\) are some natural number which represent some words ordinal value.

Answer 3

what does *occur at least as late * mean?

Answer 4

ok how about let (x,y) = (cat, batman). Is it in R?

Answer 5

I see a Cartesian product

Answer 6

Yes, it's a Cartesian product with a condition whose meaning I don't quite understand

Answer 7

at least as late means it is the same word or it comes after the word in alphabetical order.

Answer 8

\(
A\times A
\) is the set of any two words, such at \((\text{dog},\text{cat})\).

Answer 9

\(
(\text{dog},\text{cat})\notin R
\) but \(
(\text{cat},\text{dog})\in R
\) also \(
(\text{cat},\text{cat})\in R
\)

Answer 10

The second word must be the same or comes after the first word, so shouldn't (cat,dog) be in R instead?

Answer 11

What? \((\text{dog},\text{cat})\in A\times A\)

Answer 12

oh wait nvm. I just repeated what you said earlier.

Answer 13

So (cat, batman) isn't in R, but (batman, cat) is right?

Answer 14

\[A \times B =[(a,b) : a \in A \land b \in B]\] there are elements in A and B it's an ordered pair...

Answer 15

so I see two A's won't that be elements in A and elements in A again?!

Answer 16

That's correct.

Answer 17

so if A = set of all english words then we have the set of all english words and the set of all english words... like mouse mouse in A

Answer 18

the components need not to be the same. (mouse, cat) is certainly in AxA

Answer 19

hmm late for y ... like the ending of the alphabet worm xray yarn zebra

Answer 20

(digimon, pokemon)

Answer 21

(UsukiDoll, Wio) :DD

Answer 22

anyway, thanks for the explanation @Wio