- #1

- 96

- 3

I can't understand a proof in the Theorem 2.27, part (a).

If [itex]X[/itex] is a metric space, [itex]E[/itex] a subspace of X, and [itex]E'[/itex] the set of all limit points of [itex]E[/itex], we denote by [itex]\bar{E}[/itex] the set: [itex]\bar{E}=E \cup E'[/itex]

We need to prove that [itex]\bar{E}[/itex] is a closed set. Rudin's proof is this:

If [itex]x\in X[/itex] and [itex]x \notin \bar{E}[/itex] then [itex]p[/itex] is neither a point of [itex]E[/itex] nor a limit point. Hence, [itex]p[/itex] has a neighborhood which does not intersect [itex]E[/itex]. Therefore, the complement of [itex]\bar{E}[/itex] is closed.

My question is this: How do we prove that there exists a neighborhood that does not contain any limit points of [itex]E[/itex]?

If [itex]X[/itex] is a metric space, [itex]E[/itex] a subspace of X, and [itex]E'[/itex] the set of all limit points of [itex]E[/itex], we denote by [itex]\bar{E}[/itex] the set: [itex]\bar{E}=E \cup E'[/itex]

We need to prove that [itex]\bar{E}[/itex] is a closed set. Rudin's proof is this:

If [itex]x\in X[/itex] and [itex]x \notin \bar{E}[/itex] then [itex]p[/itex] is neither a point of [itex]E[/itex] nor a limit point. Hence, [itex]p[/itex] has a neighborhood which does not intersect [itex]E[/itex]. Therefore, the complement of [itex]\bar{E}[/itex] is closed.

My question is this: How do we prove that there exists a neighborhood that does not contain any limit points of [itex]E[/itex]?

Last edited: