Short answer
The support vectors are those points for which the Lagrange multipliers are not zero (there is more than just $b$ in a Support Vector Machine).
Long answer
Hard Margin
For a simple hard-margin SVM, we have to solve the following minimisation problem:
$$\min_{\boldsymbol{w}, b} \frac{1}{2} \|\boldsymbol{w}\|^2$$
subject to
$$\forall i : y_i (\boldsymbol{w} \cdot \boldsymbol{x}_i + b) - 1 \geq 0$$
The solution can be found with help of Lagrange multipliers $\alpha_i$.
In the process of minimising the Lagrange function, it can be found that $$\boldsymbol{w} = \sum_i \alpha_i y_i \boldsymbol{x}_i.$$ Therefore, $\boldsymbol{w}$ only depends on those samples for which $\alpha_i \neq 0$.
Additionally, the Karush-Kuhn-Tucker conditions require that the solution satisfies
$$\alpha_i (y_i (\boldsymbol{w} \cdot \boldsymbol{x}_i + b) - 1) = 0.$$
In order to compute $b$, the constraint for sample $i$ must be tight, i.e. $\alpha_i > 0$, so that $y_i (\boldsymbol{w} \cdot \boldsymbol{x}_i + b) - 1 = 0$. Hence, $b$ depends only on those samples for which $\alpha_i > 0$.
Therefore, we can conclude that the solution depends on all samples for which $\alpha_i > 0$.
Soft Margin
For the C-SVM, which seems to be known as soft-margin SVM, the minimisation problem is given by:
$$\min_{\boldsymbol{w}, b} \frac{1}{2} \|\boldsymbol{w}\|^2 + C \sum_i \xi_i$$
subject to
$$\forall i : \begin{aligned}y_i (\boldsymbol{w} \cdot \boldsymbol{x}_i + b) - 1 + \xi_i & \geq 0 \\ \xi_i &\geq 0\end{aligned}$$
Using Lagrange multipliers $\alpha_i$ and $\lambda_i = (C - \alpha_i)$, the weights are (again) given by $$\boldsymbol{w} = \sum_i \alpha_i y_i \boldsymbol{x}_i,$$
and therefore $\boldsymbol{w}$ does depends only on samples for which $\alpha_i \neq 0$.
Due to the Karush-Kuhn-Tucker conditions, the solution must satisfy
$$\begin{align}
\alpha_i (y_i (\boldsymbol{w} \cdot \boldsymbol{x}_i + b) - 1 + \xi_i) & = 0 \\
(C - \alpha_i) \xi_i & = 0,
\end{align}$$
which allows to compute $b$ if $\alpha_i > 0$ and $\xi_i = 0$. If both constraints are tight, i.e. $\alpha_i < C$, $\xi_i$ must be zero. Therefore, $b$ depends on those samples for which $0 < \alpha_i < C$.
Therefore, we can conclude that the solution depends on all samples for which $\alpha_i > 0$. After all, $\boldsymbol{w}$ still depends on those samples for which $\alpha_i = C$.
