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My instructor gave us an exercise to do, to show that this equality holds: $$\bigcup_{k=1}^m A_k'=\left(\bigcup_{k=1}^m A_k\right)^{\!\prime}.$$

My thoughts on approaching this question is by contradiction, suppose that there is an element that satisfies the left side of the equality and not the right side. Then there is a deleted ball with radius $r$ such that $B(x,r)$ intersect the left side of the equality is non empty while there is a deleted ball with radius $r$ such that $B(x,r)$ intersect the right side of the equality is empty. Is this the right approach in proving this? This is as far as I can go.

Thanks for the help in advance.

dfeuer
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tamefoxes
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3 Answers3

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Hint: Try doing it with two sets and then proceed by induction.

Detail on reverse inclusion: Suppose $x\notin A'\cup B'$. Then there is an open set $U$ about $x$ such that $A\cap U\subseteq\{x\}$ and there is an open set $V$ about $x$ such that $B\cap V\subseteq\{x\}$. Then $U\cap V$ is an open set about $x$ and $(U\cap V)\cap (A\cup B)\subseteq \{x\}$, so $x\notin (A\cup B)'$.

dfeuer
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It’s very easy to show that $A_\ell'\subseteq\left(\bigcup_{k=1}^mA_k\right)'$ for each $\ell\in\{1,\ldots,m\}$ and hence that

$$\bigcup_{k=1}^mA_k'\subseteq\left(\bigcup_{k=1}^mA_k\right)'\;;$$

the harder part is showing that

$$\left(\bigcup_{k=1}^mA_k\right)'\subseteq\bigcup_{k=1}^mA_k'\;.\tag{1}$$

Proof by contradiction is a reasonable guess, but it’s not needed.

Suppose that $x\in\left(\bigcup_{k=1}^mA_k\right)'$; then for each $n\in\Bbb Z^+$ there is an $$x_n\in B\left(x,\frac1n\right)\cap\bigcup_{k=1}^mA_k\;,$$ and it’s clear that the sequence $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $x$. For each $n\in\Bbb Z^+$ there is a $k_n\in\{1,\ldots,m\}$ such that $x_n\in A_{k_n}$.

  • Show that there is an $\ell\in\{1,\ldots,m\}$ such that $\{n\in\Bbb Z^+:k_n=\ell\}$ is infinite. Let $L=\{n\in\Bbb Z^+:k_n=\ell\}$.
  • Show that $\langle n:n\in L\rangle$ is a sequence in $A_\ell$ converging to $x$.
  • Conclude that $(1)$ must hold.
Brian M. Scott
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    Contradiction is still the better choice in this case. – Troy Woo Feb 03 '15 at 14:30
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    @Troy: It makes very little difference: one ends up doing pretty much the same thing either way. – Brian M. Scott Feb 03 '15 at 19:32
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    I apologize for my late follow-up, but may I ask, the mentioned contradiction could appear in the part: there is an $\ell\in{1,\ldots,m}$ s. t. $L={n\in\Bbb Z^+:k_n=\ell}$ is infinite ? If, $\forall\ell\in{1,\ldots,m}, L$ would be finite, then the there would be only finitely many terms of the sequence $(x_n)n$ in $\bigcup{k=1}^mA_k,$ which is impossible, right? – Matcha Latte Oct 30 '21 at 10:34
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    @Invisible: Not quite. For $\ell\in{1,\ldots,m}$ let $L_\ell={n\in\Bbb Z^+:k_n=\ell}$; clearly $$\bigcup_{\ell=1}^mL_\ell=\Bbb Z^+,.$$ If each $L_\ell$ were finite, then $\Bbb Z^+$ would be finite, which is impossible. That contradiction proves the first bullet point. – Brian M. Scott Oct 30 '21 at 20:12
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    Oh, I see, thank you for your kind reply, but doesn't $x\in\left(\bigcup_{k=1}^m A_k\right)'$ imply even the stronger statement $x\in B\left(x,\frac1n\right)\cap\cup_{k=1}^mA_k\color{red}{\setminus{x}}$? Then we would also be able to construct a sequence $(x_n)_n$ with infinitely many different terms and force the contradiction I mentioned? – Matcha Latte Oct 30 '21 at 22:10
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    @Invisible: You’re welcome. Yes, you can argue that for each $n\in\Bbb Z^+$ there is an $x_n\in B\left(x,\frac1n\right)\cap\bigcup_{k=1}^m\left(A_k\setminus{x}\right)$, but then you have to prove that $\langle x_n:n\in\Bbb Z^+\rangle$ has infinitely many distinct terms, and then you have to use something like my argument for the first bullet point to show that this sequence has an infinite subsequence lying in one of the sets $A_k$. Thus, you actually end up doing more work. – Brian M. Scott Oct 31 '21 at 06:24
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    @BrianM.Scott can we generalize this result to countable union? – Meet Patel Apr 20 '23 at 15:21
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    @MeetPatel: No. Working in $\Bbb R$, let $A_n=\left{\frac1n\right}$ for $n\in\Bbb Z^+$. Then $A_n'=\varnothing$ for each $n\in\Bbb Z^+$, but $\left(\bigcup\limits_{n\in\Bbb Z^+}A_n\right)'={0}$. – Brian M. Scott Apr 20 '23 at 21:38
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A slightly different way to prove

$\left(\bigcup_k A_k\right)'\subseteq \bigcup_k A_k'$

is as followed.

Say $p\in\left(\bigcup_k A_k'\right)^c$, i.e. $p$ is not a limit point of any of $A_k$, there exists a neighborhood $N(p)$ (a ball of radius $r$ centered at $p$, if you will) that does not intersect any of $A_k$.

This means $N(p)$ does not intersect $\bigcup_k A_k$ either, so therefore, $p$ cannot be a limit point of $\bigcup_k A_k$, i.e. $p\in\left(\left(\bigcup_k A_k\right)'\right)^c$.

As a result,

$\left(\bigcup_k A_k'\right)^c \subseteq \left(\left(\bigcup_k A_k\right)'\right)^c,$

so

$\left(\bigcup_k A_k\right)' \subseteq \bigcup_k A_k'$

Huy Le
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