11. Given $A=\{1,2,3, \cdots, 104\}, S$ is a subset of $A$. If $x \in S$, and both $x-1 \in S$ and $x+1 \in S$, then $x$ is called an isolated point of $S$. Then the number of all 5-element subsets of $A$ with no isolated points is $\qquad$.
11. 10000

Since the 5-element subset $S$ of $A$ has no isolated points, $S$ can only have the following 3 cases.
(1) $S=\{a, a+1, a+2, a+3, a+4\}$, where $1 \leqslant a \leqslant 100$. The number of such $S$ is 100.
(2) $S=\{a, a+1, a+2, b, b+1\}$, where $1 \leqslant a<b-3 \leqslant 100$. The number of such $S$ is $C_{100}^{2}$.
(3) $S=\{a, a+1, b, b+1, b+2\}$, where $1 \leqslant a<b-2 \leqslant 100$. The number of such $S$ is $C_{100}^{2}$.
Therefore, the number of 5-element subsets with no isolated points is $100+C_{100}^{2}+C_{100}^{2}=10000$.