TITLE: Uniqueness of analytic continuation in rigid analytic geometry
QUESTION [10 upvotes]: 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!

REPLY [3 votes]: 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.