Local property of ring of regular function If $X$ is a projective variety (irreducible algebraic subset of a projective space), $U$ is an open subset of $X$, $x\in U\subset X$, is that true that $\mathcal{O}_{X,x}=\mathcal{O}_{U,x}$ ? 
I think it is true but I do not know how to prove it. My idea is following: $\mathcal{O}_{X,x}=\lim_{x\in V}\mathcal{O}(V)$, so we can take the intersection with the given $U$ above and then taking the limit, we get the result $\mathcal{O}_{U,x}$. But I am not sure about my argument. 
Please help me to correct my idea and prove/disprove that proposition. 
 A: BenjaLim's answer is great and constitutes a direct proof of the claim in which you're interested (+1). An alternate way of looking at it is via direct limits as Brad suggests. On the one hand, $O_{X,x}=\lim_{x\in V} O(V)$ and on the other hand, $O_{U,x}=\lim_{x\in V\subseteq U} O(V)$. Can you use the universal property of direct limits to construct inverse maps $O_{X,x}\to O_{U,x}$ and $O_{U,x}\to O_{X,x}$? Hint: You've got a natural (in $V$) family of maps $O(V)\to O(U\cap V)$ for $V\ni x$ that defines $O_{X,x}\to O_{U,x}$ and conversely (the easy part, I suppose!), you've got a natural family of maps $O(V)\to O(V)$ for $U\supseteq V\ni x$ that defines $O_{U,x}\to O_{X,x}$.
Alternatively, if you're familiar with the basic theory of direct limits, just note that the index set for the direct limit defining $O_{U,x}$ is cofinal in the index set for the direct limit defining $O_{X,x}$ (according to the definitions I've given above). The proof of this is equivalent to the one suggested in the first paragraph if you unwind the logic. I'm happy to provide you with relevant definitions/more details if you'd like!
I hope this helps!
A: An element in $\mathcal{O}_{X,x}$ is an equivalence class of pairs $\langle U,s \rangle$ with $s$ a section of $\mathcal{O}_X(U)$ and $U$ open about $x$. The equivalence relation that we have is that two pairs $\langle U,s\rangle = \langle V,t\rangle$ if $s$ is "eventually equal" to $t$. That is we can find an open neighbourhood $W \subseteq U \cap V$ for which $s|_W = t|_W$. From this it is clear that $\mathcal{O}_{X,x} = \mathcal{O}_{U,x}$ because any pair on the left $\langle V,s \rangle $ is equal to a pair $ \langle U \cap V,s|_{U \cap V} \rangle$  which is an element of $\mathcal{O}_{U,x}$. Note $U \cap V \neq \emptyset$ because $x$ is in both sets. The other containment is also clear because anything open in $U$ about $x$ is also open in $X$ about $x$, since being open is a transitive property.
