In this paper, we consider tolerances induced by irredundant coverings. Each tolerance $R$ on $U$ determines a quasiorder $\lesssim_R$ by setting $x \lesssim_R y$ if and only if $R(x) \subseteq R(y)$. We prove that for a tolerance $R$ induced by a covering $\mathcal{H}$ of $U$, the covering $\mathcal{H}$ is irredundant if and only if the quasiordered set $(U, \lesssim_R)$ is bounded by minimal elements and the tolerance $R$ coincides with the product ${\gtrsim_R} \circ {\lesssim_R}$. We also show that in such a case $\mathcal{H} = \{ {\uparrow}m \mid \text{$m$ is minimal in $(U,\lesssim_R)$} \}$, and for each minimal $m$, we have $R(m) = {\uparrow} m$. Additionally, this irredundant covering $\mathcal{H}$ inducing $R$ consists of some blocks of the tolerance $R$. We give necessary and sufficient conditions under which $\mathcal{H}$ and the set of $R$-blocks coincide. These results are established by applying the notion of Helly numbers of quasiordered sets.

Thanks. We have received your report. If we find this content to be in
violation of our guidelines,
we will remove it.

Ok