Clearly the question in the title has a positive answer for analytic (or smooth, or continuous ...) curves, but what about the algebraic category? More specifically, given an irreducible polynomial curve $X \subseteq \mathbb{CP}^2$ and points $P_1, \ldots, P_k$ on $X$, when can we find another curve $Y$ (defined by a polynomial) such that the $Y$ intersects $X$ only at $P_1, \ldots, P_k$?

I find the question to be nontrivial even for $k = 1$. Here are some observations for $k = 1$ case:

If $P$ is a point on $X$ with multiplicity $\deg X - 1$, then a tangent of $X$ through $P$ intersects $X$ only at $P$ (by Bezout's theorem).

If $X$ is a rational curve and $X \setminus \{P\} \cong \mathbb{C}$, then there is a curve $Y$ such that $X \cap Y = \{P\}$.

Let $X$ be a non-singular cubic. Give it a group structure such that the origin is an inflection point. Then for all $P \in X$, there exists $Y$ such that $Y \cap X = \{P\}$ iff $P$ is a torsion point in the group.

If $X$ (of degree $d$) is non-singular at $P$, then the most direct approach for finding a $Y$ of degree $e$ intersecting $X$ only at $P$ seemed to blow it up $de$ times and look for the conditions under which $Y$ goes through each of the points on $X$ in the $i$-th infinitesimal neighborhood of $P$, $0 \leq i \leq de - 1$. But the conditions on the coefficients of the polynomial defining $Y$ did not appear very tractable.

* Edit: * I would like to make a correction to observation 3. This is what I know about a non-singular cubic curve $X$: If $P$ is an inflection point, then there is a curve $Y$ such that $Y \cap X = P$ (take $Y$ to be the tangent of $X$ at $P$). If $P$ is a non-torsion point (for the group structure on $X$ for which the origin is an inflection point), then there is no such $Y$. I don't know what happens for torsion points.

`$H^0(\mathbb{P}^2, \mathcal{O}(d)) \to H^0(X, \mathcal{O}(d)|_X)$`

. (continued) – David Speyer Nov 30 '10 at 18:21