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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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1  
Original question on MO. –  Zev Chonoles Oct 23 '11 at 19:37
    
For $m<n$ do we know that $A(n)\subseteq A(m)$ or something similar? –  Asaf Karagila Oct 23 '11 at 19:38
    
@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 :-) –  Asaf Karagila Oct 23 '11 at 19:39
    
@Asaf: I agree; I just wanted to make a note of it, not admonish Jose. –  Zev Chonoles Oct 23 '11 at 19:40
1  
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. –  Josué Tonelli-Cueto Oct 23 '11 at 20:07

2 Answers 2

up vote 3 down vote accepted

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. –  Jose Antonio Martin H Oct 23 '11 at 23:10

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. –  Jose Antonio Martin H Oct 23 '11 at 23:12
    
@Jose: So I did not prove that $P=NP$? phew, that's a load off my mind :-) –  Asaf Karagila Oct 23 '11 at 23:19
    
Umm, it seems that the problem could come from $A = \bigcup_{n=1}^{\infty} A(n).$ –  Jose Antonio Martin H Oct 24 '11 at 11:40

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