Is an open ball neighborhood equivalent to a set $V$ such as $p\in V°$ (modern definition of neighborhood) ? Stuck in the proof

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According to Rudin, Principles of Mathematical Analysis, a neighborhood of $p\in (X, \rho)$ is simply an open ball centered at $p$, that is $ \mathfrak B_{\epsilon}(p) = \{ x\in X | \rho (x,p) < \epsilon) \}$ for some $\epsilon > 0$.

According to a more modern definition a neighborhood is a set $V$ such that $p \in V°$ where $V°$ is for example defined as the greatest open set contained in $V$.

I am trying to decide whether this two definitions are equivalent in a metric space (($X,\rho)$).


  • Assume a neighborhood of $p$ is a set $V$ such that $p\in V°$.

Then, $\mathfrak B_{\epsilon}(p) = \mathfrak B_{\epsilon}(p) °$ contains $p$.

  • Now, assume a neighborhood of $p$ is a set of the form $\mathfrak B_{\epsilon}(p)$.

Then, one can clearly see that a set $V$ such that $p\in V°$ cannot be of the form $\mathfrak B_{\epsilon}(p)$ in general. (Note that we can find an $\mathfrak B_{\epsilon}(p)$ that contain $p$ however).


I don't really understand:

  • if the two definitions are not equivalent then how can Rudin establish all his analysis results without any problem using his $\mathfrak B_{\epsilon}(p)$ definition ?
  • If the two definitions are equivalent, I don't know how to go any further in my above attempt of proof. How can a set $V$ such as $p\in V°$ be an open ball centered at $p$ in general ?

Thank you very much in advance.