  Given an $n$-vertex bipartite graph $I=(S,U,E)$, the goal of set cover
problem is to find a minimum sized subset of $S$ such that every vertex in $U$
is adjacent to some vertex of this subset. It is NP-hard to approximate set
cover to within a $(1-o(1))\ln n$ factor. If we use the size of the optimum
solution $k$ as the parameter, then it can be solved in $n^{k+o(1)}$ time. A
natural question is: can we approximate set cover to within an $o(\ln n)$
factor in $n^{k-\epsilon}$ time?
  In a recent breakthrough result, Karthik, Laekhanukit and Manurangsi showed
that assuming the Strong Exponential Time Hypothesis (SETH), for any computable
function $f$, no $f(k)\cdot n^{k-\epsilon}$-time algorithm can approximate set
cover to a factor below $(\log n)^{\frac{1}{poly(k,e(\epsilon))}}$ for some
function $e$.
  This paper presents a simple gap-producing reduction which, given a set cover
instance $I=(S,U,E)$ and two integers $k<h\le (1-o(1))\sqrt[k]{\log
|S|/\log\log |S|}$, outputs a new set cover instance $I'=(S,U',E')$ with
$|U'|=|U|^{h^k}|S|^{O(1)}$ in $|U|^{h^k}\cdot |S|^{O(1)}$ time such that:
  (1) if $I$ has a $k$-sized solution, then so does $I'$;
  (2) if $I$ has no $k$-sized solution, then every solution of $I'$ must
contain at least $h$ vertices.
  Setting $h=(1-o(1))\sqrt[k]{\log |S|/\log\log |S|}$, we show that assuming
SETH, for any computable function $f$, no $f(k)\cdot n^{k-\epsilon}$-time
algorithm can distinguish between a set cover instance with $k$-sized solution
and one whose minimum solution size is at least $(1-o(1))\cdot
\sqrt[k]{\frac{\log n}{\log\log n}}$. This improves the result of Karthik,
Laekhanukit and Manurangsi.
