If I am not mistaken there is a theorem that says any curve $C$ can be embedded in $\mathbf{P}^3$. What can be said about surfaces? Do we have a theorem like:



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> All surfaces can be embedded in $\mathbf{P}^{N}$ for some fixed $N$. 
> 
> 
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And if not what is the simplest counterexample? A counterexample would be an infinite sequence of surfaces: $\{S\_1, S\_2, \cdots \}$ and infinite sequence of strictly increasing positive integers: $\{n\_1, n\_2, \cdots \}$ such that $S\_i$ can not be embedded in $\mathbf{P}^{n\_i}$. 



Embedding of algebraic surfaces
All smooth projective surfaces can be embedded in $\mathbb{P}^5$ using a (linear) projection. This is a classical theorem in algebraic geometry. In general, it is known that any smooth projective variety of dimension $n$ can be embedded into $\mathbb{P}^{2n+1}$. 


The proof runs as follows: The secant variety $Sec(X)$ (the closure of points on secants on $X$) and the tangent variety $Tan(X)$ (the closure of the union of tangent spaces) both have dimension $\le 2n+1$, so if $X$ lies in a big projective space $\mathbb{P}^N$, successive projections from points outside $Sec(X)\cup Tan(X)$, gives you a morphism $\mathbb{P}^{2n+1}$ which is injective, finite, unramified hence a closed embedding.


This result makes surfaces in $\mathbb{P}^4$ an interesting study. A remarkable theorem of Ellingsrud-Peskine says that surfaces in $\mathbb{P}^4$ are very special: those not of general type belong to finitely many families, hence one has some hope of classifying them all. For example, Hartshorne conjectured that the degree of rational surfaces in $\mathbb{P}^4$ is bounded (it is now belived that this bound is equal to 12).


There are however non-projective surfaces, which cannot be embedded in any projective space, see e.g., [this question](https://mathoverflow.net/questions/3624/nonprojective-surface).


