What is an open subset of $X$? From Walter Rudin's Principles of Mathematical Analysis Ed. 3.

2.29 Remark Suppose $E \subset Y \subset X$ where $X$ is a metric space. To say that $E$ is an open subset of $X$ means that to each point $p \in E$ there is associated a positive number $r$ such that the conditions $d(p,q) < r$, $q \in X$ imply that $q \in E$.

My understanding is that $E$ is an open subset of $X$ if for all $p \in E$, there exists $r > 0$ such that all $q \in X$ with $d(p, q) < r$ is in $E$. That is to say, every $p \in E$ has a neighborhood containing only elements of $X$ that's also in $E$.
Is this equivalent to remark 2.29?
Why state "$q \in X$ imply that $q \in E$" instead of just "$q \in E$"?
"$q \in X$ imply that $q \in E$" is true if $q \notin X$ and $q \notin E$, so the conditions are satisfied if $d(p, q) < r \wedge q \notin X \wedge q \notin E$, but this seems irrelevant when mentioning subsets of $X$.

This is exactly equivalent, Rudin is basically saying that if you are closer to $p$ than that particular radius and you are a point of $X$, then it means you are in $E$: in easier terms, it means that for every point in the open set you can always find an open ball of some tiny radius such that this ball is all contained in $E$.
Your last condition $d(p, q) < r \wedge q \notin X \wedge q \notin E$ seemes a bit a contradiction to me. No point can satisfy both the first and the second condition.
