using a bijection to prove an equality involving generating functions 
A parity-preserving subset $\{\alpha_1,\cdots, \alpha_k\}$ of $\{1,\cdots, n\}$ satisfies that $\alpha_i \cong i \mod 2$ and $\alpha_i < \alpha_{i+1}\forall i.$ Let $p_n$ be the number of parity-preserving subsets of $\{1,\cdots, n\}, n\geq 0.$ Let $P_n$ be the set of parity-preserving subsets of $\{1,\cdots, n\}$ and $D_n$ be the set of differences of $\{1,\cdots, n\},$ whose elements are the ordered pairs of the differences between consecutive elements that start with the first element of the subset. Let $\{\alpha_1,\cdots, \alpha_k\} := \sigma$ be a parity-preserving subset. Define $d(\sigma) := (\alpha_1,\alpha_2-\alpha_1,\cdots, \alpha_k - \alpha_{k-1}).$ Then $d^{-1}(\alpha_1,\alpha_2,\cdots, \alpha_k) = \{\alpha_1, \alpha_1 + \alpha_2,\cdots, \alpha_1 + \alpha_2+\cdots \alpha_k\}.$ This is called the difference-partial sum bijection. Show that $\sum_{n\geq 0} a_nx^n = \dfrac{1+x}{1-x-x^2}$ using the difference-partial sum bijection. Let $b_n$ denote the number of subsets of $\{1,\cdots, n\}$ without consecutive elements. Show that $\sum_{n\geq 0} b_nx^n = \sum_{n\geq 0} a_nx^n$ using the difference-partial sum bijection.

The two equalities should be similar to show. Given an element in $B_n$ (i.e. a subset of $\{1,\cdots, n\}$ without any consecutive elements), the difference vectors are either empty vectors or they start with a positive integer. Each subsequent difference should be at least $2$, resulting in the series $\dfrac{\frac{x}{1-x}}{1-\frac{x^2}{1-x}} = \frac{x}{1-x-x^2},$ but this is not equal to $\frac{1+x}{1-x-x^2}.$ Similarly, given an element in $P_n$, the differences are always odd (due to differing parities), which gives the series $\dfrac{\frac{x}{1-x^2}}{1-\frac{x}{1-x^2}} = \dfrac{x}{1-x-x^2},$ which is also incorrect.

What am I doing wrong? How can I account for the fact that empty sets are elements in $A_n$ and $B_n$ for all $n$ in my generating functions?

 A: If there was only one error in your solutions, it would be easy to spot, but there are actually two issues compounding one another that make it hard to find the mistake in your solution.
I will start by addressing the $b_n$ case. Your approach misses the empty subset, which obviously has no consecutive elements, however, this alone is not enough to fix the answer. Adding 1 to our generating function to account for the empty set gives $\frac{1-x^2}{1-x-x^2}$. The second error is that your approach finds the number of these subsets with maximal element n, while actually we don't care what value the biggest element is exactly, just that the biggest element is less than or equal to n. To fix this, we multiply by $\frac{1}{1-x}$ to form a partial sum of the elements, which gives the correct generating function, $\frac{1+x}{1-x-x^2}$.
As for your $p_n$ case, it is almost identical. We add in the empty subset to get $\frac{1-x^2}{1-x-x^2}$, then make sure we just bound the maximal element instead of forcing it by making a partial sum by multiplying by $\frac{1}{1-x}$, and we get the correct generating function, $\frac{1+x}{1-x-x^2}$.
