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Given two sets $A$ and $B$, I want to know the result of: $$C = A \setminus B$$

If I define the sets $A(n)$ as the set such that: $$\begin{align*} A(n) \setminus B &= \emptyset; &\quad &\forall n= 0,1,2,3,\ldots,\infty &\quad &(1)\\\ \lim_{n\to\infty} A(n) &= A &&&&(2) \end{align*}$$

If using $\lim\limits_{n\to\infty}$ over a set $A(n)$ is not possible, how can I describe the process of iterating over $n$ up to $\infty$?

Read then $\lim\limits_{n\to\infty}$ as "when $n$ approaches to $\infty$".

Then:

$$ \lim_{n\to\infty} \quad A(n) \setminus B \quad = \quad \emptyset, \quad \quad \quad \text{by} \quad (1)$$

but also,

$$\lim_{n\to\infty} \quad A(n) \setminus B \quad = \quad A \setminus B,\quad \text{by} \quad (2)$$

So, is then:

$$ A \setminus B = \emptyset{}\quad \quad?$$ That is: $$ C=\emptyset \quad? $$

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    [Original question on MO](http://mathoverflow.net/questions/78904/a-question-about-sets-closed).2011-10-23
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    For $m do we know that $A(n)\subseteq A(m)$ or something similar?2011-10-23
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    @Zev: While I agree that noting a cross posting is important, I also think that since this question was closed there it is somewhat less relevant :-)2011-10-23
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    @Asaf: I agree; I just wanted to make a note of it, not admonish Jose.2011-10-23
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    This question relies on the topology you give to the family of sets where you are working. Once you have the topology, you only have to show that complementation over every set is sequencially compact for your result, if not your result is false.2011-10-23
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    Hi, For m$n $A(n) \subseteq A(m)$ (more elements when $n$ grows) @Asaf many thanks for you help.2011-10-23
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    @Jose Antonio: And is $\lim\limits_{n\to\infty}A(n) = \cup A_n$? If not, just what kind of "limit" are you taking?2011-10-23
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    @Arturo Magidin: Yes. I agree that $\cup A_n$ is what I want to say with $\lim_{x\to\infty}$. Thank you very much to all of you for you great help.2011-10-23
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    OK, it seems that all of you converged to the same notation, arguments and conclusion.2011-10-23
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    @Jose: I really feel that you have a concrete type of sets for $A(n)$, and it would be great if you'd told us what they are.2011-10-23
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    @Jose: How do you define $A(n)$?2011-10-23
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    [link](http://arxiv.org/abs/1108.5405)2011-10-23
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    Could you add a direct reference, and not just the link? I am not inclined to read a paper in a field I am not too familiar with.2011-10-23
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    @Asaf Let me think a little bit in how to answer you more rigorously.2011-10-23
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    @Jose: I doubt I could give you any real critique on your work. I can only say that if you think that you found a simple proof to an open problem, you might want to check it again and again and then again and again, and then ask someone whose research area coincides with the paper's topic to check again and again as well.2011-10-23
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    Dear @Asaf, don't worry. I have many years on this. I am not trying to make "the revolution" nor claiming to have solved a famous problem. I'm just interested in a simple algorithm I've developed and one of the proofs I need to complete contains this set issue I ask here. This was only to let you know that this question has a special meaning and that I will apply it into a serious academical work.2011-10-23
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    Jose, while I cannot verify your argument at all, note that the union of $P_n=\{A\subseteq\mathbb N\colon |A| is countable, but $P(\mathbb N)$ is not. It might be the case here.2011-10-23
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    $A(n)$ is a parameter that defines a class of graphs, such as for instance the chromatic number $\chi(G)$. So, the union of all the classes of graphs defined by its chromatic number is the set of all graphs. So, $A(n)$ is the class of graphs with some property of size $n$.2011-10-23
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    However what is $A(\infty)$? Is it just $\bigcup A(n)$ or does it allow graphs with an infinite chromatic number? (is that even possible? I'd guess that yes but I wouldn't know since it's been years since discrete mathematics, and it was only with finite graphs anyway)2011-10-23

2 Answers 2

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You specify in comments that $n\leq m$ implies $A(n)\subseteq A(m)$. that suggests that your "limit" operation is meant to be a union; that is, when you write $\lim\limits_{n\to\infty}A(n) = A$, you are "really" saying that $$A = \bigcup_{n=1}^{\infty} A(n).$$

If this is the case, then we have: $$\begin{align*} A\setminus B &= \left(\bigcup_{n=1}^{\infty}A(n)\right)\setminus B\\ &= \bigcup_{n=1}^{\infty}(A(n)\setminus B). \end{align*}$$ To see the last equality: if $x\in (\bigcup\limits_n A(n)) \setminus B$, then there exists $n$ such that $x\in A(n)$, and $x\notin B$; in particular, there exists $n$ such that $x\in A(n)\setminus B$, so $x$ lies in $\bigcup\limits_n(A(n)\setminus B)$. Conversely, if $x\in \bigcup\limits_n(A(n)\setminus B)$, then there exists $n$ such that $x\in A(n)\setminus B$, so $x\notin B$, and $x\in A(n)\subseteq \bigcup\limits_k A(k)$. So $x\in (\bigcup\limits_n A(n)) \setminus B$, as desired.

Therefore, if $A(n)\setminus B = \emptyset$ for each $n$, then $A\setminus B = \emptyset$.

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    Thanks a lot very illustrative. Also thank you for editing the question, it now looks very good formatted.2011-10-23
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Added the information from the comment that $A(n)\subseteq A(m)$ for $n\le m$, if we have that $A=A_\infty = \bigcup \Big\{A_n\mid n\in\mathbb N\Big\}$.

We can use the following claim:

$$A\setminus B=\emptyset\iff A\subseteq B$$

Proof:

  • Suppose $A\subseteq B$ then $x\in A\setminus B$ if and only if $x\in A$ and $x\notin B$. However $x\in A\rightarrow x\in B$ therefore $A\setminus B=\varnothing$.

  • Suppose $A\nsubseteq B$, then for some $x\in A$ we have $x\notin B$ therefore for this $x$ we have $x\in A\setminus B$. Therefore $A\setminus B\neq\varnothing$.

Now we want to calculate $A\setminus B$, however $x\in A\rightarrow x\in A_n$ for at least one $n\in\mathbb N$. By the claim, $A_n\subseteq B$ therefore $x\in B$ and so $A\subseteq B$.

Again by the claim, $A\setminus B=\varnothing$.

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    Thanks for your kind help Asaf. This helped very much to me to understand my mistakes.2011-10-23
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    @Jose: So I did not prove that $P=NP$? *phew*, that's a load off my mind :-)2011-10-23
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    Umm, it seems that the problem could come from $A = \bigcup_{n=1}^{\infty} A(n).$2011-10-24