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The notation to verify if an element belongs to a set is $e \in E$. But which notation should I use to verify if an element is part of an ordered pair? Is $a \in (a,b)$ valid (e.g., $1 \in (1,2)$)?

Thanks!

user136598
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  • I don't think there is an accepted standard notation for saying a particular element appears in an ordered pair. Certainly "$1 \in (1,2)$" is not accepted notation. – Bill Cook Mar 19 '14 at 19:54
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    Although, some set theorists "construct" ordered pairs some sets as follows: $(x,y) = {x,{x,y}}$. In such a case, $1 \in (1,2)$ is true! :) – Bill Cook Mar 19 '14 at 19:55
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    @BillCook: Amusingly, though, $2 \not \in (1,2)$ by the same definition. – Ross Millikan Mar 19 '14 at 19:57
  • Thanks. In my case $V = { (a,b) | a \in A, b \in B, condition }$. Then, I need to verify if a given element is part of any $(a,b) \in V$. So $a \in (a,b)$ probably not apply and I'm not sure which notation to use. – user136598 Mar 19 '14 at 19:59
  • In that case you're just asking if you belong to ${ x \in A ;|; \exists b \in B \mbox{ such that } (x,b) \in V } \cup { x \in B ;|; \exists a \in A \mbox{ such that } (a,x) \in V }$ – Bill Cook Mar 19 '14 at 20:06
  • @RossMillikan :) – Bill Cook Mar 19 '14 at 20:18
  • Yes, I think that ${ x \in A | \exists b \in B such that (x,b) \in V}$ is exactly what I need! Thank you! :) – user136598 Mar 19 '14 at 20:19

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As I stated in my comment above, I am not aware of any accepted standard notation for saying a particular element appears in an ordered pair.

That said, suppose we take the view that order $n$-tuples are "really" maps from $\mathbb{Z}_{>0} = \{1,2,3,\dots\}$ into our set $X$. Or more generally from an ordinal into $X$. (This sort of thing is done in field such as Topology where one runs into infinite Cartesian products).

From this viewpoint: $(a,b)=f$ where $f:\{1,2\} \to X$ and $f(1)=a$, $f(2)=b$. If we take this view $a, b \in \mathrm{Range}(f)$. So we could write $1 \in \mathrm{Range}(1,2)$.

However, try doing that without explaining yourself and you'll get some odd looks. :)

Bill Cook
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  • Thanks for your answer! It certainly would solve the problem, but you are right when you say that it is not intuitive.

    Since the order in not important for me $(a,b)$ and $(b,a)$ would have the same effect maybe I could say that V is a set of sets with 2 elements where one element belongs to A and the other belongs to B. Is something like: $V = {{a,b} | a \in A, b \in B, condition}$ valid?

     – user136598 Mar 19 '14 at 20:13
  • Yes. Your $V$ will now be a set of subsets of size 2 (and 1 if $a=b$). So now instead of being a subset of $A \times B$, your $V$ will be a subset of the power set $\mathcal{P}(A \cup B)$. – Bill Cook Mar 19 '14 at 20:16
  • Then your question becomes: Does there exist some $X \in V$ such that $a \in X$. :) – Bill Cook Mar 19 '14 at 20:17
  • So, $V = {{a,b}| a \in A, b \in B, f(a,b)>5 }$ would create a set with subsets of size 2 (a and b are never equal). Then I can ask if $\exists {a,b} \in V | a' \in {a,b}$. The last question is, can I use that condition when building the set (f(a,b) could be a distance function for an example)? :) – user136598 Mar 19 '14 at 20:33
  • Sure. That's a perfectly valid condition to put in the middle of a definition. – Bill Cook Mar 19 '14 at 20:57