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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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    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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    Umm, it seems that the problem could come from $A = \bigcup_{n=1}^{\infty} A(n).$2011-10-24