Dimension of irreducible affine variety is same as any open subset Let $X$ be an irreducible affine variety. Let $U \subset X$ be a nonempty open subset. Show that dim $U=$ dim $X$.
Since $U \subset X$, dim $U \leq$ dim $X$ is immediate. I also know that the result is not true if $X$ is any irreducible topological space, so somehow the properties of an affine variety have to come in. I have tried assuming $U=X$ \ $V(f_1,...,f_k)$ but I don't know how to continue on.
Any help is appreciated!

You should note that any nonempty open subset of an irreducible variety (or topological space) is dense : https://en.wikipedia.org/wiki/Hyperconnected_space.
Then you can use the definition of the dimension to conclude.
