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Hierarchy of the IRS Classes

The strictness of IRS can be obtained by just removing the label $\cL(x)$ from $\cI(x,y)$, and therefore by increasing by one at most its compactness (or linearity compactness). Because a cyclic interval is the union of two linear intervals, we have the trivial collection of inclusions:

For every $k \geqs 1$,
\item $k$-...
...bseteq (k+1)$-$\mbox{$\cS\cL\cI\cR\cS$}_\star$.

Open question 13    
- Is 1- $\mbox{$\cS\cI\cR\cS$}= 1$- $\mbox{$\cI\cR\cS$}$?
- Same question for k- $\mbox{$\cS\cI\cR\cS$}_d$, k > 1?

In the following, $X \subsetneq Y$ means that $X \subseteq Y$ and $X
\neq Y$. Because in [FJ88] and in [BvLT91] it is shown $K_{2,2k+1} \in (k+1)$- $\mbox{$\cL\cI\cR\cS$}_d$ but not in k- $\mbox{$\cI\cR\cS$}$, we have:

For every $k \geqs 1$, $k$-$\mbox{$\cI\cR\cS$}_d \subsetneq (k+1)$-$\mbox{$\cI\cR\cS$}_d$.

For every $k \geqs 1$,
\item ...
...cR\cS$}_d \subsetneq 1$-$\mbox{$\cI\cR\cS$}_d$.

Theorem 57, and 58 can be improved by Theorem 47.

For every $k \geqs 1$,
\item $k$-$\mbox{$...
...\cS$}_d \subsetneq k$-$\mbox{$\cI\cR\cS$}_d$.

From Theorem 47, for every $k \geqs 1$, $K_{2k+1} \in
\floor{(2k+1)/2}$- $\mbox{$\cL\cI\cR\cS$}_d = k$- $\mbox{$\cL\cI\cR\cS$}_d$, and also $K_{2k+1} \in
k$- $\mbox{$\cS\cI\cR\cS$}_d$. From the second point of Theorem 47, $K_{2k+1}
\in t$- $\mbox{$\cS\cL\cI\cR\cS$}_d$ implies that $t \geqs \ceil{(2k+1)/2} = k+1$. Hence $K_{2k+1} \notin k$- $\mbox{$\cS\cL\cI\cR\cS$}_d$. It follows that k- $\mbox{$\cS\cL\cI\cR\cS$}_d \neq
k$- $\mbox{$\cL\cI\cR\cS$}_d$ and k- $\mbox{$\cS\cL\cI\cR\cS$}_d \neq
k$- $\mbox{$\cS\cI\cR\cS$}_d$. Similarly $K_{2k+2} \in k$- $\mbox{$\cI\cR\cS$}$, but $K_{2k+2} \notin k$- $\mbox{$\cL\cI\cR\cS$}_d$, and $K_{2k+2} \notin k$- $\mbox{$\cS\cI\cR\cS$}_d$.