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I have a sequence of ordered pairs: $$(0,1), (0,2), (0,3), (0,4), \ldots, (0, n),$$ $$(1,2), (1,3), (1,4), \ldots, (1, n),$$ $$(2,3), (2,4), \ldots, (2, n),$$ $$(3,4), \ldots, (3, n),$$ $$\ldots,$$ $$(m,n),$$ where $m<n$ $\forall m, n \in \mathbb{N}_0$ and the maximal value of $m$ equals $n-1$.

Question. How to write down this sequence with the set notation?

Edit. My attempt is: $A=\{(m,n): m<n, max(m)=n-1, m,n \in \mathbb{N}_0 \}$.

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  • $\begingroup$ $m<n$ and $m=n+1$? $\endgroup$
    – Arnaud Mortier
    May 4, 2018 at 0:27
  • $\begingroup$ Also, in your attempt $m$ and $n$ are dummy variables: they can't be, they are fixed by the problem. $\endgroup$
    – Arnaud Mortier
    May 4, 2018 at 0:28
  • $\begingroup$ @ArnaudMortier, $m$ is less than $n$ but the maximal value of $m$ equals $n+1$. $\endgroup$
    – Nick
    May 4, 2018 at 0:31
  • $\begingroup$ So $n+1<n$?${}$ $\endgroup$
    – Arnaud Mortier
    May 4, 2018 at 0:33
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    $\begingroup$ @Nick I think you mean that the maximal value of $m$ is $n-1$. Otherwise it contradicts that $m < n$. So for $n = 5$ you can have $(0,5), (1,5), (2,5), (3,5)$ and $(4,5)$ but not $(5,5), (6,5),\ldots$ If that is the case then the fact that $n,m\in \mathbb{N}_0$ and $m < n$ already encodes that the maximal value for $m$ given a fixed $n$ is $n-1$, so you can simply write $$A = \{(m,n)\ :\ m < n,\ m,n\in \mathbb{N}_0\}$$ or simply $$A = \{(m,n)\in \mathbb{N}_0^2,\ m < n\}.$$ $\endgroup$
    – Darth Geek
    May 4, 2018 at 0:34

1 Answer 1

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I suspect that what you want is : $A_n=\{(i,j)\in\mathbb N^2\mid 0\le i\le n-1,\ i+1\le j\le n\}$

where the last couple is $(n-1,n)$ and $m=n-1$

Can be shortened to $A_n=\{(i,j)\in\mathbb N^2\mid 1\le i+1\le j\le n\}$

Note that $A_n$ is finite and depends on the value of $n$.

In your attempt, since you put $n\in\mathbb N$ inside the set, then $A$ become infinite and is just equal to $\mathbb N\times\mathbb N^*$. I'm not sure this is what you expected.

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