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Let $(X, d)$ be a metric space, $E \subseteq X$ and $x \in X \setminus E$. Prove that the following are equivalent:

  1. $x \in \overline E$
  2. $x \in \operatorname{Der}(E) = \{x \text{ is an accumulation point of } E\}$
  3. every open neighborhood of $x$ contains an infinite number of points in $E$
  4. there exists a sequence $\{x_n\} \subseteq E$ such that $\lim\limits_{n \to +\infty} x_n = x$

I already proved $1) \implies 2)$ and $2) \implies 3)$. I know how to do $4) \implies 1)$. It's $3) \implies 4)$ that is giving me problems. I really don't see how a neighborhood having infinite points in $E$ would guarantee that such a sequence exists. Could you give me a hint?

I know that I could also prove $3) \implies 2)$, $2) \implies 1)$ and $1) \iff 4)$ and be done with it, but that's a lot of work and I'm really curious about how $3)$ relates to $4)$.

rubik
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1 Answers1

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Hint for 3 implies 4: Let $n$ be a positive integer and $x_n\in E\cap B(x,1/n)$, $x_n$ converges towards $x$.

Tsemo Aristide
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  • Ok, I see how that works. For this to work it's necessary that $B(x, 1)$ contains an infinite number of points in $E$, which is exactly what $3)$ guarantees. – rubik Apr 30 '16 at 16:26
  • @rubik No, you need only 1 point in every $B(x,\frac{1}{n}) \cap E$, which is what 1 already garantuees. – Henno Brandsma May 01 '16 at 11:11
  • @HennoBrandsma Well that's equivalent I think. Every $B(x, \frac1n) \cap E$ has $1$ point only if $B(x, 1)$ has infinite. Because if $B(x, 1)$ had a finite number of point there would be a $n$ such that $B(x, \frac1n) \cap E$ has no points. – rubik May 01 '16 at 19:52