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:
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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).
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If $X$ is a rational curve and $X \setminus \{P\} \cong \mathbb{C}$, then there is a curve $Y$ such that $X \cap Y = \{P\}$.
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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.
Best Answer
Here are some considerations on the case $X$ smooth.
Let $d$ be the degree of $X$ and let $L$ be the restriction to $X$ of $O_{P^2}(1)$.
If $k=1$ then the condition is precisely that the line bundle $L(-dP)$ is a torsion point of $Pic^0(X)$. In fact let $m$ be such that $mL(-dP)$ is trivial. Since the map $H^0(P^2,{\cal O}_{P^2}(m))\to H^0(X, mL)$ is onto, there exists a curve $Y$ of degree $m$ that intersects $X$ precisely at $P$ with multiplicity $md$. So the condition is satisfied for at most countably many points $P\in X$, unless $X$ is rational. One can argue in a similar (more complicated) way for $k>1$.
I don't know if the remark that follows is useful. If $X$ is smooth of genus $g$, $P\in X$ is fixed and $k=g+1$, then one can consider the image of the map $X^g\to Pic^0(X)$ defined by mapping $(P_1,...,P_g)$ to $(g+1)L(-d(P+P_1+...+P_g))$. This map is surjective, so the above argument implies that, given $P$, one can find $g$ points such that there exists a curve $Y$ that intersects $X$ only at $P, P_1,\dots P_{g}$. Since the subvarieties {P_i=P} of $X^g$ and the weak diagonal map to proper subvarieties of $Pic^0(X)$ and the torsion points are dense in $Pic^0(X)$, one can find $P,P_1,...P_g$ distinct. More generally, if $k>g$ one can assign $k-g$ points and find a curve $Y$ that meets $X$ at those points and at precisely $g$ additional points.