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I would like to write $\{(i, x, y) \mid (i,x) \in \{(1,1), (2,1), (2,2)\}\}$ where $y$ could be any (integer) value. Is there a shorter way to write that...

Also for $\{(i, x, y) \mid i = 2, x = 1\}$, can we write $\{(2, 1, y)\}$ which looks like a singleton?

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4 Answers

up vote 2 down vote accepted

$$\{(i, x, y) \mid y \in \mathbb{Z}, (i,x) \in \{(1,1), (2,1), (2,2)\} \}$$

For second one

$$ \{ \hspace{3pt} (2, 1, y) \mid \hspace{3pt} y \in \mathbb{Z} \}$$

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Not exceptionally better, but your set is

$$\{1,2\} \times \{1,2\} \times \mathbb{Z}$$

The set $\{(2,1,y) | y \in \mathbb{Z}\}$ is pretty compact notation as it is.

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The first proposition is wrong. –  Did Mar 3 '12 at 20:34
    
Ah, true enough. I didn't notice that $(1,2,y)$ is not valid. –  Kurtis Zimmerman Mar 3 '12 at 21:05
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The second is easy: $\{2\} \times \{1\} \times \mathbb{Z}$. For the first, you could write $\{(1,2),(2,1),(2,2)\}\times \mathbb{Z}$ if you're willing to identify $((i,x),y)$ with $(i,x,y)$.

Or for the second: $\{(2,1,y):\;y\in\mathbb{Z}\}$

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For your first set, you could use the notation for cartesian products, and write $\{(1, 1), (2, 1), (2, 2)\} \times \mathbb{Z}$. (If you interpret this literally, its elements look like $((2, 2), 42)$, so you might want to note that you're identifying this with $(2, 2, 42)$.)

Likewise, for your second set, you can write $\{(2, 1)\} \times \mathbb{Z}$. Alternatively, you can abuse notation a bit, and write $(2, 1) \times \mathbb{Z}$, or $(2, 1, \mathbb{Z})$.

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