Elegant notation?

1. Sep 3, 2006

bomba923

Is the notation
$$\left\{ \begin{gathered} {\text{inequality}} \hfill \\ {\text{inequality}} \hfill \\ {\text{inequality}} \hfill \\ \end{gathered} \right\}$$
generally accepted to denote the solution set (i.e., intersection) of the inequalities?

If so, then the following (part of a problem I came up with) should be easy to understand:
$$\bigcup\limits_{\begin{subarray}{l} \left( {i,j,k} \right) \in \mathbb{N}^3 , \\ i < j < k \leqslant n \end{subarray}} {\left\{ \begin{gathered} \left( {y - y_i } \right)\left( {x_i - x_j } \right)\left| {\begin{array}{*{20}c} {x_i - x_k } & {y_i - y_k } \\ {x_i - x_j } & {y_i - y_j } \\ \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_j } \right)\left| {\begin{array}{*{20}c} {x_i - x_k } & {y_i - y_k } \\ {x_i - x_j } & {y_i - y_j } \\ \end{array} } \right| \hfill \\ \left( {y - y_i } \right)\left( {x_i - x_k } \right)\left| {\begin{array}{*{20}c} {x_i - x_j } & {y_i - y_j } \\ {x_i - x_k } & {y_i - y_k } \\ \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_k } \right)\left| {\begin{array}{*{20}c} {x_i - x_j } & {y_i - y_j } \\ {x_i - x_k } & {y_i - y_k } \\ \end{array} } \right| \hfill \\ \left( {y - y_j } \right)\left( {x_j - x_k } \right)\left| {\begin{array}{*{20}c} {x_j - x_i } & {y_j - y_i } \\ {x_j - x_k } & {y_j - y_k } \\ \end{array} } \right| \geqslant \left( {x - x_j } \right)\left( {y_j - y_k } \right)\left| {\begin{array}{*{20}c} {x_j - x_i } & {y_j - y_i } \\ {x_j - x_k } & {y_j - y_k } \\ \end{array} } \right| \hfill \\ \end{gathered} \right\}}$$

Is there a simpler/more elegant way to express this, or is it fine the way it is?

Last edited: Sep 3, 2006
2. Sep 3, 2006

HallsofIvy

That looks like a set of inequalities rather than the solution set to me. Rather like saying that {x2- 4= 0} denotes the set {2, -2}. (The second way is more "elegant"!)

3. Sep 3, 2006

bomba923

Well, to be more clear, perhaps I should add intersection symbols:
$$\bigcup\limits_{\begin{subarray}{l} \left( {i,j,k} \right) \in \mathbb{N}^3 , \\ i < j < k \leqslant n \end{subarray}} {\left( \begin{gathered} \left( {y - y_i } \right)\left( {x_i - x_j } \right)\left| {\begin{array}{*{20}c} {x_i - x_k } & {y_i - y_k } \\ {x_i - x_j } & {y_i - y_j } \\ \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_j } \right)\left| {\begin{array}{*{20}c} {x_i - x_k } & {y_i - y_k } \\ {x_i - x_j } & {y_i - y_j } \\ \end{array} } \right| \cap \hfill \\ \left( {y - y_i } \right)\left( {x_i - x_k } \right)\left| {\begin{array}{*{20}c} {x_i - x_j } & {y_i - y_j } \\ {x_i - x_k } & {y_i - y_k } \\ \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_k } \right)\left| {\begin{array}{*{20}c} {x_i - x_j } & {y_i - y_j } \\ {x_i - x_k } & {y_i - y_k } \\ \end{array} } \right| \cap \hfill \\ \left( {y - y_j } \right)\left( {x_j - x_k } \right)\left| {\begin{array}{*{20}c} {x_j - x_i } & {y_j - y_i } \\ {x_j - x_k } & {y_j - y_k } \\ \end{array} } \right| \geqslant \left( {x - x_j } \right)\left( {y_j - y_k } \right)\left| {\begin{array}{*{20}c} {x_j - x_i } & {y_j - y_i } \\ {x_j - x_k } & {y_j - y_k } \\ \end{array} } \right| \hfill \\ \end{gathered} \right)}$$
Essentially, I wish to denote a union of solution sets :shy:
Would that be clear/understood from the way I rewrote it here?

Last edited: Sep 3, 2006
4. Sep 3, 2006

shmoe

It looks like you are still thinking of an inequality as the set of things that satisfy it, this isn't the usual use of the notation. The set of x's that satisfy x>2 would be written:

{x| x>2}

(specifiy some set x is coming from if it's not clear from the context, like x a real number, or integer), not

x>2

5. Sep 3, 2006

bomba923

Thanks shmoe
In that case, this would be the more correct way to denote my union of solution sets:
$$\bigcup\limits_{\begin{subarray}{l} \left( {i,j,k} \right) \in \mathbb{N}^3 , \\ i < j < k \leqslant n \end{subarray}} {\left\{ {\left( {x,y} \right)\left| \begin{gathered} \left( {y - y_i } \right)\left( {x_i - x_j } \right)\left| {\begin{array}{*{20}c} {x_i - x_k } & {y_i - y_k } \\ {x_i - x_j } & {y_i - y_j } \\ \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_j } \right)\left| {\begin{array}{*{20}c} {x_i - x_k } & {y_i - y_k } \\ {x_i - x_j } & {y_i - y_j } \\ \end{array} } \right| \hfill \\ \left( {y - y_i } \right)\left( {x_i - x_k } \right)\left| {\begin{array}{*{20}c} {x_i - x_j } & {y_i - y_j } \\ {x_i - x_k } & {y_i - y_k } \\ \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_k } \right)\left| {\begin{array}{*{20}c} {x_i - x_j } & {y_i - y_j } \\ {x_i - x_k } & {y_i - y_k } \\ \end{array} } \right| \hfill \\ \left( {y - y_j } \right)\left( {x_j - x_k } \right)\left| {\begin{array}{*{20}c} {x_j - x_i } & {y_j - y_i } \\ {x_j - x_k } & {y_j - y_k } \\ \end{array} } \right| \geqslant \left( {x - x_j } \right)\left( {y_j - y_k } \right)\left| {\begin{array}{*{20}c} {x_j - x_i } & {y_j - y_i } \\ {x_j - x_k } & {y_j - y_k } \\ \end{array} } \right| \hfill \\ \end{gathered} \right.} \right\}}$$
Right?

Last edited: Sep 3, 2006
6. Sep 3, 2006

shmoe

Sure, I'd call that more correct, but you'd probably want to put some "or"'s in there.

edit-spelling, I wasn't calling you "Sue"

Last edited: Sep 3, 2006
7. Sep 3, 2006

bomba923

Some "or's" ? Why so?

8. Sep 3, 2006

shmoe

maybe "and"'s then. You just have a list of inequalities as conditions,

{x|x>2, x>4, x=65}

What does this mean? It's ambiguous as it's written.

9. Sep 3, 2006

bomba923

Do you mean like this:
$$\bigcup\limits_{\begin{subarray}{l} \left( {i,j,k} \right) \in \mathbb{N}^3 , \\ i < j < k \leqslant n \end{subarray}} {\left\{ {\left( {x,y} \right)\left| \begin{gathered} \left( {y - y_i } \right)\left( {x_i - x_j } \right)\left| {\begin{array}{*{20}c} {x_i - x_k } & {y_i - y_k } \\ {x_i - x_j } & {y_i - y_j } \\ \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_j } \right)\left| {\begin{array}{*{20}c} {x_i - x_k } & {y_i - y_k } \\ {x_i - x_j } & {y_i - y_j } \\ \end{array} } \right| \wedge \hfill \\ \left( {y - y_i } \right)\left( {x_i - x_k } \right)\left| {\begin{array}{*{20}c} {x_i - x_j } & {y_i - y_j } \\ {x_i - x_k } & {y_i - y_k } \\ \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_k } \right)\left| {\begin{array}{*{20}c} {x_i - x_j } & {y_i - y_j } \\ {x_i - x_k } & {y_i - y_k } \\ \end{array} } \right| \wedge \hfill \\ \left( {y - y_j } \right)\left( {x_j - x_k } \right)\left| {\begin{array}{*{20}c} {x_j - x_i } & {y_j - y_i } \\ {x_j - x_k } & {y_j - y_k } \\ \end{array} } \right| \geqslant \left( {x - x_j } \right)\left( {y_j - y_k } \right)\left| {\begin{array}{*{20}c} {x_j - x_i } & {y_j - y_i } \\ {x_j - x_k } & {y_j - y_k } \\ \end{array} } \right| \hfill \\ \end{gathered} \right.} \right\}}$$
Correct?

10. Sep 3, 2006

AKG

Yes, post #9 is correct.