I have the following maybe trivial quesion regarding real algbraic sets:
Suppose $V\subset \mathbb{R}^n$ is an irreducible real algebraic variety (i.e. it can not be written as union of two algebric varieties) and let $W\subset V$ be an algebraic variety. Does $\dim W=\dim V$ imply $W=V$?
I know that this is true in for complex algebraic sets, but I also know that in real analytic setting many things can fail (hopefully not for algebraic).
I belive that the answer is positive but I've been looking at the literature Bochnak et al., Cox et al., and Shafarevich and I couldn't found a definitve answer. So if the ansewer is positve then please give hint of an argument, or else provide an counterexample. (Ps: I'm mostly interested in the case where $\dim X=1$).
If you're looking just on the level of points, then yes, this is true. If an irreducible topological space $X$ is a closed subspace of an irreducible topological space $Y$ then the dimension of $Y$ must be at least the dimension of $X$, with equality iff $X=Y$: any chain of proper inclusions of closed irreducible subsets of $X$ is also a chain of proper inclusions of closed irreducible subsets of $Y$, and if $X$ is a proper closed subset of $Y$ then we can always extend any chain of proper closed subsets by plopping $Y$ on the end.