I have been goind through Rudin for the past month and I have arrived at this theorem in the topology chapter. Now I understand Rudin's proof but I was trying to come up with my own and I am not sure if it is correct or not.

Theorem: Every neighbourhood is an open set.

Proof: Suppose that there exists a neighbourhood $N$ of $p$ that is not open. Then there exists a point $q$ in $N$ that is not an interior point of $N$. Then any neighbourhood $N_q$ of $q$ is not included in $N$. Choose $N_q$ such that $p$ is in $N_q$. Since $N_q$ is not in $N$, then $p$ is not in in $N$. Contradiction.

I have attached an image showing what I had in mind when I wrote this.

The gray edge boundary means that it's not included

  • $\begingroup$ What is your definition of a neighbourhood? $\endgroup$ – user2520938 Sep 10 '17 at 19:16
  • $\begingroup$ A neighbourhood of a point p is a set N consisting of all points q such that d(p, q) < r, where r>0. $\endgroup$ – user3535525 Sep 10 '17 at 19:21
  • $\begingroup$ Oke, then what is your definition of an open set? $\endgroup$ – user2520938 Sep 10 '17 at 19:21
  • $\begingroup$ A set is open if every point of that set is an interior point of that set. And a point p is interior of E if there is a neighbourhood N of p such that N is in E. $\endgroup$ – user3535525 Sep 10 '17 at 19:23
  • $\begingroup$ You should probably include this in the question, since they are equivalent but non-standard definitions. $\endgroup$ – user2520938 Sep 10 '17 at 19:25

I think your proof is incorrect. There is a problem with your statement: "Since $N_q$ is not in $N$, then $p$ is not in $N$".

It is true that $N_q$ is not a subset of $N$, but it is certainly possible that some elements (points) of $N_q$ are also elements of $N$.

  • $\begingroup$ You are right, I see my error now. Thank you. $\endgroup$ – user3535525 Sep 10 '17 at 19:28

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