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Rudin makes the following definitions:

(a) A neighborhood of p is a set $N_r(p)$ consisting of all $q$ such that $d(p, q) < r$, for some $r > 0$.

(b) $E$ is open if every point of $E$ is an interior point of $E$.

(c) $p$ is an interior point of $E$ if there exists a neighborhood $N_r(p)$ of $p$ that is a subset of $E$.

This is my attempt at the proof that every neighborhood $N = N_r(p)$ is an open set:

Let $x \in N$. Then there exists a neighborhood of $x$ that is also a subset of $N$, namely $N$ itself. Since $x$ and $N$ were arbitrary, every neighborhood is an open set.

Rudin's proof involves the use of the metric, and I wonder why it wouldn't use a more general approach, or if my proof is at all correct.

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    $\begingroup$ I thought a neighborhood of $x$ was a set $V$ such that there exists an open set $U$ with $x\in U\subset V$. It does not have to be a neighborhood of all its points, so it does not have to be open. $\endgroup$
    – almagest
    Commented Jun 24, 2016 at 16:53

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Given Rudin's definition of "neighborhood," if $x\in N_r(p)$ but $x\not=p$, then $N_r(p)$ is not, itself, a neighborhood of $x$. Strictly speaking, a set that is a neighborhood of a point is a ball with that point as its center, so $N_r(p)$ can only be a neighborhood of $p$. What Rudin therefore needs to prove -- and where metric considerations enter in -- is that if $x\in N_r(p)$, then there is some ball centered at $x$, $N_{r'}(x)$, that is completely contained in $N_r(p)$.

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    $\begingroup$ This was helpful, so in essence all that is necessary is show that there exists a neighborhood around $x \in N_r(p), x \neq p$ that is a subset of $N_r(p)$, which is clear if we make $r' = r - d(p, x)$. $\endgroup$
    – lazopard
    Commented Jun 24, 2016 at 17:28
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    $\begingroup$ @lazopard, sounds good. Depending on the level of rigor that's called for, you may need to mumble something about the triangle inequality, to prove that $N_{r-d(p,x)}(x)\subset N_r(p)$. $\endgroup$ Commented Jun 24, 2016 at 17:55
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In his Principles of Mathematical Analysis, Rudin defines "neighborhoods" as what we commonly call "open balls", and denoted $B(x,r)$ (Rudin also adopts this terminology in his Real and Complex Analysis). The standard definition of a neighborhood $N$ of $x$ is a set with an open subset $V$ such that $$ x\in V \subset N. $$ A more restrictive definition would be that $N$ is an open set containing $x$. These two definitions almost always serve the same purpose.

As for your proof, you need to show that $B(x,r)$ contains open balls about other points in it. When you say "$x$ and $N$ are arbitrary", it doesn't mean other arbitrary $N$s fall in your current $B(x,r)$.

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The problem with your proof is that you're assuming that $N$ is a "neighborhood" of $x$ for every $x\in N$. That is not actually the case with the (slightly non-standard) definitions you quote.

For example, on the real line, $N=(1,5)=N_2(3)$ is a "neighborhood" of $3$, and $2\in N$, but $(1,5)$ is not a "neighborhood" of $2$. So you cannot, given these definition, use $N$ itself as a witness that $2$ is an interior point of $N$.

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You can only define the neighborhood of a point or a subset so the fact you want to show does not make sense. N is a neighborhood of p if there exists an open subset which contains p and which is contained in N.

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