Let $\mathcal{L}_{d}^{n}$ be the $\binom{d+n}{n}-1$ dimensional projective space of hypersurfaces of degree $d$ in $\mathbb{P}^{n}$ and $p_{1},\ldots,p_{r}\in\mathbb{P}^{n}$. We denote by $\mathcal{L}_{d}^{n}(p_{1},\ldots,p_{r})\subseteq \mathcal{L}_{d}^{n}$ the projective space of hypersurfaces of degree $d$ in $\mathbb{P}^{n}$ that passes through $p_{1},\ldots,p_{r}$.
$\mathcal{L}_{d}^{n}(p_{1})$ is a hyperplane of $\mathcal{L}_{d}^{n}$, i.e. it is a projective subspace given by a linear restriction. Similarly, $\mathcal{L}_{d}^{n}(p_{1},\ldots,p_{r})$ is a projective subspace of $\mathcal{L}_{d}^{n}$ given by $r$ linear restrictions.
We say that $p_{1},\ldots,p_{r}$ are in $d$-general position, if the restrictions imposed on $\mathcal{L}_{d}^{n}$ by $\binom{d+n}{n}$ or fewer of them are linearly independent.
I want to prove that if $p_{1},\ldots,p_{r}\in\mathbb{P}^{n}$ are in $d$-general position, then they are in $1$-general position. I do not know if it is true, but I think so.
The answer is no. Consider $3$ distinct points in $\mathbb{P}^2$ that lie on a line. Clearly they are not in $1$-general position. But they are in $2$-general position since you can always find a curve of degree two that goes through two of the points but not through all three of them.