Talk:Maximum-minimums identity

This identity may be stated thus:

\max\{x_1,x_2,\ldots,x_{n}\} = \sum_{\varnothing \neq A \subseteq \{\,1,\dots,n\,\}} (-1)^{\left|A\right|+1} \min A. $$

But notice the caveat: A is not empty. Now a case could be made, based on vacuous truth and similar to the case for empty sums and empty products, for regarding the minimum of the empty set as the same thing as the maximum of the full set {1, ..., n}:
 * $$ \min\varnothing = \max\{x_1,x_2,\ldots,x_{n}\}.\, $$
 * $$ \min\varnothing = \max\{x_1,x_2,\ldots,x_{n}\}.\, $$

If that is done, then we can drop that caveat and state the identity like this:

\sum_{A \subseteq \{\,1,\dots,n\,\}} (-1)^{\left|A\right|+1} \min A = 0. $$

(And then we could let the exponent be |A| instead of |A| + 1; maybe that could be regarded as making the whole thing just a bit simpler.) Michael Hardy (talk) 21:56, 5 September 2008 (UTC)