example of a dissipated space is given by:

\[lemma-lots-dis\] If $X$ is a compact LOTS, then $X$ is dissipated

The proof (see Lemma 3.4 of [@KU2]) shows that given $g \in {{\mathfrak M}}(X)$, there is a finer $f \in {{\mathfrak M}}(X)$ such that $f(X)$ is a compact metric LOTS and $f$ is order-preserving.

Note that just having *one* tight map $g$ from $X$ onto some metric compactum $Z$ is not sufficient to prove that $X$ is dissipated, since the tightness of $g$ says nothing at all about the complexity of a particular $g{^{-1}}\{z\}$. However, if all $g{^{-1}}\{z\}$ are scattered, then just one tight $g$ is enough by Lemma 3.5 of [@KU2]:

\[lemma-one-enough\] Assume that some $g \in {{\mathfrak M}_{\mathfrak S}}(X)$ is $\kappa$–tight. Then all $f \le g$ are also $\kappa$–tight, so that $X$ is $\kappa$–dissipated.

This suggests the following definition:

\[def-super\] $\pi\in{{\mathfrak M}}(X)$ is $\kappa$–*supertight* iff $\pi$ is $\kappa$–tight and $\pi\in{{\mathfrak M}_{\mathfrak S}}(X)$. Then $X$ is $\kappa$–*superdissipated* iff some $\pi\in{{\mathfrak M}_{\mathfrak S}}(X)$ is $\kappa$–supertight.

Using Lemmas \[lemma-one-enough\], \[lemma-local\], and \[lemma-MS-down\] above:

\[lemma-lim-finer\] If $\pi, \sigma \in{{\mathfrak M}_{\mathfrak S}}(X)$, $\pi \le \sigma$, and $\sigma$ is $\kappa$–supertight, then $\pi$ is $\kappa$–supertight.

A compactum $X$ is $\kappa$–superdissipated iff $X$ is $\kappa$–dissipated and ${{\mathfrak 