Example 4 Let $X_{n}=\{1,2, \cdots, n\}$, for any subset $A$ of $X_{n}$, denote $t(A)$ as the smallest element in $A$. Find $t=$ $\sum_{A \subseteq X_{n}} t(A)$.
Explanation: Classification of the values of $t(A)$:
Let $t(A)=i(i=1,2, \cdots, n)$, then
$$
\sum_{1 \leq X_{n}} t(A)=\sum_{i=1}^{n} \sum_{t(A)=i, A \subseteq X_{n}} i \text {. }
$$

When $t(A)=i$, $A$ does not contain any number smaller than $i$, and the $n-i$ numbers larger than $i$ can either belong to $A$ or not, each having two possibilities, so there are $2^{n-i}$ such subsets $A$, meaning $i$ appears $2^{n-i}$ times in the sum $\sum_{t(A)=i, A \subseteq X_{n}} i$. Therefore,
$$
\begin{array}{l}
t=\sum_{1 \leq X_{n}} t(A)=\sum_{i=1}^{n} \sum_{t(A)=i, A \subseteq X_{n}} i=\sum_{i=1}^{n} i \times 2^{n-i}, \\
2 t=\sum_{i=1}^{n} i \times 2^{n-i+1}=\sum_{i=0}^{n-1}(i+1) \times 2^{n-i} \\
=2^{n}+\sum_{i=1}^{n-1}(i+1) \times 2^{n-i} \\
=2^{n}+\sum_{i=1}^{n-1} i \times 2^{n-i}+\sum_{i=1}^{n-1} 2^{n-i} .
\end{array}
$$

Subtracting the two equations gives
$$
t=2^{n}+\sum_{i=1}^{n-1} 2^{n-i}-n=2^{n+1}-2-n .
$$