# A Zariski open subset of a variety has the same dimension as the variety.

I am reading Joe Harris' book Algebraic Geometry: A first course. In the book he says:

The Grassmannian $G(k,n)$ contains, as a Zariski open subset, affine space $\mathbb{A}^{k(n-k)}$, and thus has dimension $k(n-k)$.

Is it true that a Zariski open subset of a variety has the same dimension as the variety? I am trying to find it in Harris' book, but I do not get it. Is it an inmediate consequence of any other result?

I would appreciate if you could explain this fact to me, or if you could tell me a book where it is shown.

Dimension of the variety $X$ is the transcendence degree of its function field $k(X)$ which is isomorphic to $k(U)$ for an open set $U \subset X$ so the dimension consequence.