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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?

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  • $\begingroup$ 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
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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 $$ A = \bigcup_{x\in\mathbb R, d\in(0,1]} \{ (x,x+d+n) \mid n\in\mathbb N \} $$

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  • $\begingroup$ 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. $\endgroup$ – Henning Makholm 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 $n\mapsto \sqrt n\sin(\log n)$. $\endgroup$ – Henning Makholm 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

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