Can't understand a proof in Rudin

  • Thread starter kostas230
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  • #1
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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]?
 
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  • #2
verty
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If every open ball at x contains a limit point of E, x would be a limit point of those limit points. Can you prove that it is contradictory for an open ball at x to be disjoint from E?

Edit: I changed "neighborhood" to "open ball", I think neighborhood is wrong although Rudin may define it differently.
 
  • #3
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I found a more straightforward proof. Suppose [itex]p[/itex] is a limit point of [itex]E'[/itex]. Then, for every neighborhood [itex]N_r (p)[/itex] there exists a limit point [itex]q[/itex] of [itex]E[/itex]. There exists a real number [itex]h>0[/itex] with [itex]d(p,q)=r-h[/itex]. Consider the neighborhood [itex]N_h (q)[/itex]

[itex]d(s,p) \leqslant d(p,q)+d(q,s) < r[/itex]

Hence, [itex]s\in N_r (p)[/itex]. Therefore, [itex]p[/itex] is a limit point of [itex]E[/itex].
Did I get it right? xD
(BTW, I should mention that I'm a physics undergrad; no training in rigorous math except a linear algebra self study with Axler. So, I'm a little slow in this. xD)
 
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  • #4
verty
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That works. It is more technical than what I was alluding to above but it's fine.

You should probably mention that s is a point in E.
 
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