A specific partition of the interval $[1,n]$ of $\mathbb{N}$

Question

Let $n>0$. I would like to partition the interval $[1,n]$ of $\mathbb{N}$ in a specific manner. For example, if $n=6$, the partition I want is $$[1,2]\cup[3,4]\cup[5,6].$$ Is there a general way to write this partition for arbitrary $n$? If $([x,y])_{(x,y)\in[1,n]\times[1,n]}$ denotes this partition, then I know that the $x$-component will have to run through the odd integers of the interval and the $y$-component will run through the even integers. But I would like a cleaner description. How does one specify this construction in general?

Answer 1

For something like this, I would just use an iterated union notation. Mostly something like this just tends to involve pattern recognition. You know your intervals will have the form $[2i \pm 1, 2i]$ (use the sign of your desire but it'll affect the bounds), because of their formatting and the numbers' parity. But owing to this, you can't naively just throw in $n$ as the upper bound, since it'd get doubled by the operand, so you'll need to halve it.

For instance, you could use

$$\bigcup_{i=1}^{n/2} [2i-1,2i]$$

but only on the premise $n$ is an even integer and ensured to be sure. If not, then you'll run into a weird problem with the last set, namely because you'll have an odd number hanging out all alone. To remedy this, you could do the following involving the floor function:

$$\{n\} \cup \bigcup_{i=1}^{\lfloor n/2 \rfloor} [2i-1,2i]$$

(or otherwise handle that leftover integer $n$ however you please).

Answer 2

Your set is not a partition of $[1,n]$, but you can write it like this: $$\bigcup_{i=1}^{n/2} [2i-1,2i]$$

Update after your edit: If you want only integers, this union just reduces to $\{1,2\,\dots,n\}$.

Maybe you want a set of ordered pairs instead? $$\{(2i-1,2i)\}_{i=1}^{n/2}$$