In my Finite-Element Method class, we defined a meshing $\tau_{h}$ for an interval $I = [0,10]$ with $h$ being the size of a cell (unit in the partition of $I$). Then, internal approximation requires a finite dimensional subspace of the bigger function space $V$ in which the solution of a PDE lies.
Consider the partition of $I$ into $[x_{i}, x_{i+1}]_{i=1}^{n}$. One example of the approximating space is $ V_{h, 1} = \{ v \in \mathscr{C}^{0}(I), v_{[x_{i}, x_{i+1}]} \in \mathbb{P}_{1}[X], j = 0, ...N \text{ with } v(0)=v(10)=0 \} $.
In the notes and books that I'ved consulted, the dimension of $V_{h,1}$ is claimed to be $N$. I am unable to understand why. My (erroneous) reasoning was that if we have to pick coefficients $a_{i} + b_{i}x$ for each cell $[x_{i}, x_{i+1}]$, then the dimension should be $2N$, which is wrong according to the book.
Further into the chapter, there is a description for an approximating space of $\mathbb{P}_{2}[X]$ polynomials: $V_{h, 2} = \{ v \in \mathscr{C}^{0}(I), v_{[x_{i}, x_{i+1}]} \in \mathbb{P}_{2}[X], j = 0, ...N \text{ with } v(0)=v(10)=0 \}$.
The book claims that this space has dimensions $2N+1$ but I am unable to understand why.
How does one go about calculating the dimension of an approximating/meshing space?