Algebraic curve approximation I am wondering wether it exists a theorem that any continuous path on the plane one can
approximate with algebraic curve $P(x,y)=0$ ($P$- is a polynom)?
 A: Given any compact set $K$ in the plane (in particular the image of a compact interval under a continuous function) and $\epsilon > 0$, there is a finite set $\{(x_j, y_j)\}_{j=1}^n \subseteq K$ such that $K$ is contained in the union of the disks of radius $\epsilon$ centred at $(x_j, y_j)$.  Then $K$ is within distance $\epsilon$ in the Hausdorff metric
of the real algebraic curve $P(x,y) = 0$, where $P(x,y) = \prod_{j=1}^n ((x-x_j)^2 + (y-y_j)^2 - \epsilon^2)$. 
A: It's hard to tell what this question means exactly. 
If a "continuous path" is a continuous image of the unit interval, then any continuous path can be uniformly approximated by polynomial paths; this is the Weierstrass approximation theorem.  Any of those polynomial paths can be extended to a polynomial image of the entire real line, which (because you are in the plane) is a set-theoretic complete intersection (i.e. the locus of $P(x,y)=0$ for some $P$).  
But --- the continuous path consisting of a line segment in the plane is of course not the vanishing set of any polynomial, nor does there seem to be any reasonable sense in which it could be approximated by such.
On the other hand, if a "continuous path" is a continuous image of the real line, then it can be uniformly approximated on compact sets by polynomial paths, each of which is the locus of vanishing of some $P$.  Whether this satisfies your needs depends on what you mean by approximating the path.  
A: There might be problems with the Peano curve which is a continuous map $[0,1]\to\mathbb{R}^2$ whose image is the square $[0,1]\times [0,1]$.
