0
$\begingroup$

I am trying to represent the following set as the countable union of sets

$A = \{(x,y)|x<y\}$

I know that $A = \bigcup \{\{x|x<a\} \times \{y|a<y\}\}$ where $a$ is a real number

My problem is that I need to represent set $A$ using the countable union of sets using the cartesian product (as shown above), and in particular I need variable $a$ to be a function of natural numbers not reals. For example, $a$ can be $1/n$ where $n$ is a natural number ( this 1/n is wrong but I am just adding it for clarification).

Is such representation using natural numbers possible?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ Your set builder notation for A is wrong because of mishandling of a. $\endgroup$ – William Elliot May 17 '18 at 21:21
  • $\begingroup$ Um.... 1 is a countable number. A union of 1 set is a countable union. $\endgroup$ – fleablood May 17 '18 at 22:29
2
$\begingroup$

You can let your $a$ range over $\mathbb Q$ instead of $\mathbb R$.

If $x$ and $y$ are real numbers such that $x<y$, there is always a rational between them, so $(x,y)$ will be in one of your sets.

Hopefully you already know that $\mathbb Q$ is countable.

However, what that gives you is not $A$ as a union of countable sets, but $A$ as a countable union of sets that that are themselves not countable.

For an actual union of countable sets, there has to be uncountably many of those sets. And for that I don't think we can get better than something silly like $$ A = \bigcup_{x\in\mathbb R, d\in(0,1]} \{ (x,x+d+n) \mid n\in\mathbb N \} $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your reply, I will edit my question, what I need actually is the countable union of the sets in the form of the cartesian product of two sets using a natural number, I know that it can work with rational numbers, but I want it using natural numbers $\endgroup$ – user123 May 17 '18 at 21:45
  • $\begingroup$ @user123: Do you not know that the rationals are countable? You can index your sets by $\mathbb N$ and apply your favorite surjection $\mathbb N\to\mathbb Q$ in the definition of each set instead. $\endgroup$ – hmakholm left over Monica May 17 '18 at 21:50
  • $\begingroup$ I know that rationals are countable, but is there a way to use natural numbers directly without going through rationals? $\endgroup$ – user123 May 17 '18 at 21:54
  • $\begingroup$ @user123: Pick any function with domain $\mathbb N$ whose range is dense in $\mathbb R$. One possibility with a short expression would be $n\mapsto \sqrt n\sin(\log n)$. $\endgroup$ – hmakholm left over Monica May 17 '18 at 21:56
  • $\begingroup$ OK, regarding your first comment above "You can index your sets by N and apply your favorite surjection N→Q in the definition of each set instead", how can I do this? and can I do this with real not rationals? $\endgroup$ – user123 May 17 '18 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.