Uniqueness of analytic continuation in rigid analytic geometry In classical complex analysis it is easy to prove that a meromorphic function has at most one analytic continuation (on an open connected subset of $\mathbb C$, say).
The problem of non-uniqueness of analytic continuation is one of the reasons why it is not possible (if one wants a good theory) to translate the complex theory to the $p$-adic case without some modifications, and so it is one of the motivation for introducing rigid analytic variety. However, I am not able to find a precise statement that explains under which hypothesis the uniqueness of analytic continuation holds for rigid analytic varieties. So the question is the following.
Let $k$ be a non-archimedean field and let $X$ be a connected rigid analytic space over $k$. Let $f \colon X \to k$ be a rigid analytic function that vanishes on $Y$, that is an admissible subdomain of $X$. It is always true that $f$ vanishes on $X$? If this is not the case, under which assumptions it is true?
Any references will be very appreciated!
 A: No: let $X$ be the union of the coordinate axes in the affine plane.  As over $\mathbf{C}$, the answer is affirmative on a connected normal analytic space. Hint: prove in any rigid-analytic space that connected components are witnessed via finite linked chains of connected affinoid opens (and after thereby reducing to the affinoid case, assume smoothness if you are unfamiliar with the excellence properties of affinoid algebras). 
The answer is applicable to meromorphic functions as well, but proving that requires more care (e.g., one has to first figure out how to appropriately define the concept of meromorphicity and prove some basic features of it).
A: Let me just complement xbnv's answer with a mild generalization.  If $X$ is an irreducible rigid space (and let's suppose we're dealing with reduced spaces from the outset), then its normalization $\tilde{X}$ is connected and normal and is equipped with a finite surjective map $\tilde{X}\to X$.  Using xbnv's answer, it follows that the statement holds for $X$ as well.
In short, if you replace "connected" by "irreducible" then you're good.
