The dimension of an irreducible scheme $X$ at $x$, dim$_x(X)$ is defined as the smallest dimension among its open neighbourhoods and the dimension of its local ring dim$(\mathcal{O}_{X,x})$ is just the Krull dimension.
There are many properties of those two dimensions, as well as the Krull dimension of a scheme dim$X$, and their relationships. For example, dim$X$=sup$_x$ dim$_x(X)$=dim$(\mathcal{O}_{X,x})$ for closed points $x$.
I am wondering
(1) if it is true that dim$_x(X)$=dim$(\mathcal{O}_{X,x})$ for a closed point $x$?
(2) And how to prove $\dim _ x(X) \geq \dim (\mathcal{O}_{X, x})$ in general? (I guess it is true...)
Thank you all in advance!