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If p is a limit point of a set E, then every neighborhood of p contains infinitely many points of E

Proof: Suppose there is a neighborhood N of p which contains only a finite number of points of E. Let r be the minimum of the distances of these points from p. The minimum of a finite set of positive numbers is clearly positive so that r > 0. The neighborhood contains no point q of E such that q not equal to p which contradicts the fact that p is a limit point of E.

What somehow just jumps over my head is : how is it clear that the neighborhood contains no point q in E s.t. q not equal to p? I thought we'd just been talking about a finite list of point say q1 .... qn which are all not equal to p?

Juniper Leaf
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    I think "the neighborhood" in the last sentence is supposed to be a smaller neighborhood, defined using $r$. (Otherwise your objection applies, and also $r$ is not used, which is suspicious.) – Trevor Wilson Sep 29 '13 at 20:54
  • I'd basically completely misread this proof. I missed out where he called the first neighborhood N and the second one Nr. – Juniper Leaf Sep 29 '13 at 21:02

1 Answers1

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There are only finitely many points in your neighborhood $N$. Now, these points are not $p$, so $\delta_i=d(p,q_i)>0$ for each $i=1,\ldots,r$ say. Now, if we look at $\min \delta_i=\delta$, the neighborhood $B(p,\delta)$ contains none of the $q_i$, and it is contained in $N$. If I recall correctly, Rudin uses the term neighborhood precisely for open balls, so you may assume $N=B(p,r)$ for some $r>0$. If $N$ is an open set containing $p$ and finitely many of $q_i$ the proof is the same, since having found $\delta$ you can simply choose $\delta'<\delta$ until $B(p,\delta')\subseteq N$, which will happen for $N$ is open.

ADD Indeed, Rudin says: "A neighborhood of $p$ is a set $N_p(r)$ consisting of all $q$ such that $d(p,q)<r$."

We may call $x$ a limit point of a set $A$ if every punctured neighborhood $N$ of $x$ has nonempty intersection with $A$, or by saying every neighborhood of $N$ contains some point of $A$ distinct from $x$. We may also call $x$ an accumulation point of $A$ if every neighborhood $N$ of $x$ has infinite intersection with $A$, that is, if every nbhd $N$ of $x$ contains infinitely many points of $A$. One can readily see that if $x$ is an accumulation point of $A$ the it is a limit point of $A$. Rudin has proven the converse, thus showing these two definitions coincide in metric spaces. A slight generalization can be done: the concepts of accumulation point and limit point conincide whenever $(X,\mathcal T)$ is Hausdorff, and the proof is essentially the same, where we now separate the $q_i$s from $p$ by open sets.

Pedro
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