Elegant Notation: Intersection of Inequalities

bomba923 · Sep 3, 2006

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?

HallsofIvy · Sep 3, 2006

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"!)

bomba923 · Sep 3, 2006

HallsofIvy said:

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"!)

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?

shmoe · Sep 3, 2006

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

bomba923 · Sep 3, 2006

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?

shmoe · Sep 3, 2006

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"

bomba923 · Sep 3, 2006

shmoe said:

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

Some "or's" ? Why so?

shmoe · Sep 3, 2006

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.

bomba923 · Sep 3, 2006

shmoe said:

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.

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?

AKG · Sep 3, 2006

Yes, post #9 is correct.

Elegant Notation: Intersection of Inequalities

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