1
$\begingroup$

Considering a finite sequence of real numbers $x_0, x_1, ..., x_n$, which has no particular order, could I write a single statement that says the difference between any two elements of the sequence is less than some real number $y$?

In other words, can I write all of the following inequalities as a single statement?

$|x_0 - x_1| < y, |x_0 - x_2| < y, ..., |x_0 - x_n| < y, |x_1 - x_2| < y, |x_1 - x_3| < y,...,|x_1 - x_n| < y,.., |x_{n-1} - x_n| < y$

EDIT: More specifically, I'm looking for a single inequality, involving $y$ and some linear combination of $x_0, x_1, ..., x_n$, that is only true if the above list of inequalities is true.

$\endgroup$
2
$\begingroup$

Sure, how about "$|x_i-x_j|<y$ for all $i$ and $j$". Or if you want to be more formal; $$(\forall i,j\in\{0,\ldots,n\})(|x_i-x_j|<y).$$ Alternatively, you could write $\max_{i,j}\{|x_i-x_j|\}<y$.

$\endgroup$
  • $\begingroup$ This made me realize that I'm actually looking for an inequality involving $y$ and some linear combination of $x_0, x_1, ..., x_n$ that has the same meaning as the list of inequalities. I'll update the question now. $\endgroup$ – lthompson Mar 10 at 1:40
  • 1
    $\begingroup$ @lthompson I have added an alternative. There is no single inequality of the form $$\left|\sum_{i=0}^nc_ix_i\right|<y,$$ that is equivalent with the above, if that is what you are hoping for. $\endgroup$ – Servaes Mar 10 at 1:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.