A simple case of Euclidean space vector

Question

How can I see that (5.1.1) in the snippet below arrange $$\theta(x)$$ in non-increasing order ? It is clear to me for $t=1$ and $t=d$.

Answer 1

To see that $\theta$ must have a non-increasing ordering, it may be best to consider an example. For this purpose, we consider a sequence $H$ of constant real-valued functions $h_{i}$ with $i \in \{1,2,3,4,5\}$ on $X$ given by $H=(1,4,3,5,2)$.

Obviously, if $t=1$, then we have $5$ different ways to arrange a restriction on $H$ in only one coordinate. Taking the minimum of $\{1\}$ is $1$, taking the minimum of $\{4\}$ is $4$ etc. so that we have to take the maximum over the set $\{1,4,3,5,2\}$ which is $5$.

If $t=5$, then the restriction is the complete sequence $H$. Taking its minimum value at $1$, which is also its maximum value.

It remains to discuss $t=2,3,4$. For $t=2$, we consider a restriction on two possible coordinates, that can be arranged into $10$ different ways. That produces these $10$ sets

$$\{\{1, 4\}, \{1, 3\}, \{1, 5\}, \{1, 2\}, \{4, 3\}, \{4, 5\}, \{4, 2\}, \{3, 5\}, \{3, 2\}, \{5, 2\}\} $$

taking the minimum from each of these sets yields

$$ \{1, 1, 1, 1, 3, 4, 2, 3, 2, 2\}.$$

Taking from this set its maximum is $4$.

Proceeding in this way for $t=3,4$, gives for $t=3$ a maximum value of $3$ and for $t=4$ a value of $2$.

Thus, we get for $\theta$ the non-increasing ordering $(5,4,3,2,1)$.

You may want to check out by yourself the more cumbersome example

$$H=(1,4,3,5,2,15,-6,-3,-10,26,45,4,-3,26) $$

which should produce the following non-increasing order:

$$\theta=(45,26,26,15,5,4,4,3,2,1,-3,-3,-6,-10).$$