6. For an $n$-element set $H_{n}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, the number of subsets is $a_{n}=2^{n}$. For all these subsets, we design a method to conveniently arrange them in a sequence, such as the empty set $\varnothing$ being placed at position 0, $\left\{x_{1}\right\}$ at position 1, $\left\{x_{2}\right\}$ at position 2, $\left\{x_{1}, x_{2}\right\}$ at position 3, $\left\{x_{3}\right\}$ at position 4, and so on. Then, the subset at position 100 is . $\qquad$
6. $\left\{x_{3}, x_{6}, x_{7}\right\}$ Detailed explanation: First, establish a certain mathematical mapping between each element $x_{1}, x_{2}, \cdots, x_{n}$ and its index number $n$: let $x_{1}$ correspond to $2^{0}, x_{2}$ correspond to $2^{1}, \cdots, x_{n}$ correspond to $2^{n-1}$; the binary subset $\left\{x_{2}, x_{4}\right\}$ corresponds to $2^{1}+2^{3}, \cdots, the n$-element subset $\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$ corresponds to the ordinal number $2^{0}+2^{1}+\cdots+2^{n-1}$. For this reason, the subset $\left\{x_{3}\right\}$ corresponds to the ordinal number $2^{2}=4$, i.e., $\left\{x_{3}\right\}$ is in the 4th position. Conversely, the subset at the 100th position can be found: $100=2^{6}+2^{5}+2^{2}$ corresponds to the subset $\left\{x_{3}, x_{6}, x_{7}\right\}$.