Can someone please point out the mistake in the following "theorem"?

$A_n \subseteq \mathbb{R}^p$ is closed for all $n \in \mathbb{N}$ then $\displaystyle A = \bigcup_{n=1}^\infty A_n$ is closed in $\mathbb{R}^p$.

Proof: We shall show that $A^\prime \subseteq A$ where $A^\prime$ denotes the accumulation points of $A$

Choose any $x \in A^\prime$. Then $B^\prime (x,\delta)\cap A \neq \emptyset \forall \delta >0$.

$\implies B^\prime (x,\delta)\cap \bigg(\displaystyle \bigcup_{n=1}^\infty A_n \bigg) \neq \emptyset \ \ \ \ \forall \delta >0$

$\implies$ there exists a $n_0 \in \mathbb{N}$ such that $B^\prime (x,\delta) \cap A_{n_0} \neq \emptyset \ \ \ \ \forall \delta >0$.

$\implies x \in A^\prime_{n_0} \subseteq A_{n_0} \implies x \in \displaystyle \bigcup_{n=1}^\infty A_n = A$

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    $\begingroup$ What does prime mean? Closure? Boundary? Other? In case you don't care all the way for the error here, there's also an easy counterexample, let $A_n=\left[{1\over n}, 1-{1\over n}\right]$. Then the union is $(0,1)$, which is clearly not closed. $\endgroup$ – Adam Hughes Apr 7 '15 at 23:19
  • $\begingroup$ @AdamHughes - Accumulation Point $\endgroup$ – user860374 Apr 7 '15 at 23:20
  • $\begingroup$ @AdamHughes - I know about the counterexample and that the theorem is not true :). I just cannot seem to find the mistake in the given proof :) $\endgroup$ – user860374 Apr 7 '15 at 23:21
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    $\begingroup$ The error is that you assume just because the ball intersects $A_{n_0}$ that somehow implies $x\in A'_{n_0}$. You only know some point in the ball is also in $A'_{n_0}$, and as $\delta$ changes, that intersection point will naturally change as well. $\endgroup$ – Adam Hughes Apr 7 '15 at 23:26
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    $\begingroup$ The best thing you can do for yourself is to take any counterexample, e.g. $A_n=\{\frac 1n\}$, and try to see for yourself where this breaks down. It will teach you more about finding the weak spots in a proof, more than any answer on this site could. $\endgroup$ – Asaf Karagila Apr 7 '15 at 23:31

HINT: Look at $n_0$, is it the same for all $\delta>0$? Just be careful with the order of quantifiers: is it $\exists n_0\in \mathbb{N}:\forall \delta >0$ or $\forall \delta >0:\exists n_0\in \mathbb{N}$?

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  • $\begingroup$ Thank you! :). This pointed the mistake out quite clearly! :) $\endgroup$ – user860374 Apr 7 '15 at 23:28
  • $\begingroup$ You're welcome @Dillon, glad It helped. $\endgroup$ – Daniel Apr 7 '15 at 23:33

Besides $A'$ meaning closure, the notation is still bad. In a fatal way. I would have written $n_\delta$ instead of $n_0$... which would act as a warning for the quantifier sitting in front of $\delta$. In other words, the "choice" of $n$ depends on $\delta$ and you can't say "for all" in the line before the last.

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  • $\begingroup$ That's precisely, the error in the proof. $\endgroup$ – Daniel Apr 7 '15 at 23:26
  • $\begingroup$ Thank you! :). In our textbook $A^\prime$ means accumulation points and $\bar{A}$ means closure :) . $\endgroup$ – user860374 Apr 7 '15 at 23:29
  • $\begingroup$ I bet it has a problem that asks you to prove $A'=\bar A$. $\endgroup$ – Oskar Limka Apr 7 '15 at 23:41
  • $\begingroup$ @OskarLimka But $A' \neq \bar A$ in general (take a singleton subset of the reals – it is closed, but has no accumulation points). We only have $\bar A = A \cup A'$. $\endgroup$ – Eike Schulte Apr 8 '15 at 6:51
  • $\begingroup$ I see. It depends how one defines accumulation point. In Latin countries (France, Italy, Spain, etc.) you usually don't exclude the limit from the approximating sequence (or approximating neighborhoods if you prefer a topological approach). This makes an isolated point (including a singleton) an accumulation point and you avoid unnecessary distinctions. I appreciate though that Anglo-Saxon, especially North-American, and possibly German, literature has it somewhat differently. $\endgroup$ – Oskar Limka Apr 8 '15 at 13:36

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