3
$\begingroup$

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 $B^\prime (x,\delta) \cap A_{n_0} \neq \emptyset \ \ \ \ \forall \delta >0$.

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

$\endgroup$
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ 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
5
$\begingroup$

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 $\forall \delta >0:\exists n_0\in \mathbb{N}$?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.