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8

$begingroup$

There is a similar topic about finite intersections and the constructs for that case are pretty clear (for example, limits of real numbers approached by monotonically increasing sequences of rational numbers). But, is that the case for finite symmetric differences? It not obvious at all for me... What the construct can be if it is possible (is it possible to find the bijection between almost disjoint family and a family of sets with finite symmetric differences)? What's the counter-example if it's not?

elementary-set-theory

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Viktor ZayakinViktor Zayakin

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  $begingroup$
  I'm not sure why this got a vote to close.
  $endgroup$
  – Noah Schweber
  8 hours ago
  

  
  
  
  

add a comment | 

8

$begingroup$

There is a similar topic about finite intersections and the constructs for that case are pretty clear (for example, limits of real numbers approached by monotonically increasing sequences of rational numbers). But, is that the case for finite symmetric differences? It not obvious at all for me... What the construct can be if it is possible (is it possible to find the bijection between almost disjoint family and a family of sets with finite symmetric differences)? What's the counter-example if it's not?

elementary-set-theory

share|cite|improve this question

asked 8 hours ago

Viktor ZayakinViktor Zayakin

4333 bronze badges

New contributor

Viktor Zayakin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure why this got a vote to close.
  $endgroup$
  – Noah Schweber
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

There is a similar topic about finite intersections and the constructs for that case are pretty clear (for example, limits of real numbers approached by monotonically increasing sequences of rational numbers). But, is that the case for finite symmetric differences? It not obvious at all for me... What the construct can be if it is possible (is it possible to find the bijection between almost disjoint family and a family of sets with finite symmetric differences)? What's the counter-example if it's not?

elementary-set-theory

share|cite|improve this question

asked 8 hours ago

Viktor ZayakinViktor Zayakin

4333 bronze badges

New contributor

Viktor Zayakin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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asked 8 hours ago

Viktor ZayakinViktor Zayakin

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Check out our Code of Conduct.

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asked 8 hours ago

Viktor ZayakinViktor Zayakin

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asked 8 hours ago

Viktor ZayakinViktor Zayakin

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asked 8 hours ago

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11

$begingroup$

No, this is not possible.

Think about it this way: saying that $Xtriangle Y$ is "small" is saying that $X$ is "close to" $Y$. So by counting the number of sets "close to" a given one, we can get a bound on the size of a family of sets whose pairwise symmetric differences are all "small."

Specifically, suppose $XsubseteqmathbbN$ (we might as well take $mathbbN$ to be our countable "base set"). Then there is a bijection between the sets close to $X$ and the finite sets of naturals - send a finite $F$ to the close-to-$X$ set $Xtriangle F$, and to invert this send a close-to-$X$ set $Y$ to the finite set $Xtriangle Y$ - and there are only countably many of the latter.

• An important point here is that $triangle$ allows cancellation: if $Xtriangle A=Xtriangle B$ then $A=B$. So the same finite difference can't occur more than once, and $Xtriangle F$ is the only set whose symmetric difference with $X$ is $F$ (and hence the map described above is surjective).

So any collection $mathcalF$ of subsets of $mathbbN$ (or any countable set) whose pairwise symmetric differences are finite must be countable: picking some $XinmathcalF$, there are only countably many subsets of $mathbbN$ whose symmetric difference with $X$ is finite.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Noah SchweberNoah Schweber

137k1010 gold badges163163 silver badges310310 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Am I right that you mean that any finite symmetric difference (there are only countably many of them) can be repeated only countably many times by any two distinct sets that creating a sub-family of sets which have this specific symmetric difference?
  $endgroup$
  – Viktor Zayakin
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorZayakin No - it can't be repeated at all! If $Xtriangle A =X triangle B$ then $A=B$. Each finite set corresponds to exactly one set close to $X$.
  $endgroup$
  – Noah Schweber
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorZayakin But that's not what I'm doing. I'm saying fix a single element $X$ of $mathcalF$ and think about things from $X$'s perspective. Everything in $mathcalF$ has to be close to $X$, but there are only countably many things close to $X$ in the first place.
  $endgroup$
  – Noah Schweber
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, but it is possible that $mathitX triangle A = Y triangle B$ for some four sets. Sorry for using complicated wording (I'm not fluent enough). By the way, your last comment cleared it all for me, thanks a lot. This problem is done... Moving on!
  $endgroup$
  – Viktor Zayakin
  8 hours ago
  
  

  
  
  
  

add a comment | 

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11

$begingroup$

No, this is not possible.

Think about it this way: saying that $Xtriangle Y$ is "small" is saying that $X$ is "close to" $Y$. So by counting the number of sets "close to" a given one, we can get a bound on the size of a family of sets whose pairwise symmetric differences are all "small."

Specifically, suppose $XsubseteqmathbbN$ (we might as well take $mathbbN$ to be our countable "base set"). Then there is a bijection between the sets close to $X$ and the finite sets of naturals - send a finite $F$ to the close-to-$X$ set $Xtriangle F$, and to invert this send a close-to-$X$ set $Y$ to the finite set $Xtriangle Y$ - and there are only countably many of the latter.

• An important point here is that $triangle$ allows cancellation: if $Xtriangle A=Xtriangle B$ then $A=B$. So the same finite difference can't occur more than once, and $Xtriangle F$ is the only set whose symmetric difference with $X$ is $F$ (and hence the map described above is surjective).

So any collection $mathcalF$ of subsets of $mathbbN$ (or any countable set) whose pairwise symmetric differences are finite must be countable: picking some $XinmathcalF$, there are only countably many subsets of $mathbbN$ whose symmetric difference with $X$ is finite.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Noah SchweberNoah Schweber

137k1010 gold badges163163 silver badges310310 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Am I right that you mean that any finite symmetric difference (there are only countably many of them) can be repeated only countably many times by any two distinct sets that creating a sub-family of sets which have this specific symmetric difference?
  $endgroup$
  – Viktor Zayakin
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorZayakin No - it can't be repeated at all! If $Xtriangle A =X triangle B$ then $A=B$. Each finite set corresponds to exactly one set close to $X$.
  $endgroup$
  – Noah Schweber
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorZayakin But that's not what I'm doing. I'm saying fix a single element $X$ of $mathcalF$ and think about things from $X$'s perspective. Everything in $mathcalF$ has to be close to $X$, but there are only countably many things close to $X$ in the first place.
  $endgroup$
  – Noah Schweber
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, but it is possible that $mathitX triangle A = Y triangle B$ for some four sets. Sorry for using complicated wording (I'm not fluent enough). By the way, your last comment cleared it all for me, thanks a lot. This problem is done... Moving on!
  $endgroup$
  – Viktor Zayakin
  8 hours ago
  
  

  
  
  
  

add a comment | 

11

$begingroup$

No, this is not possible.

Think about it this way: saying that $Xtriangle Y$ is "small" is saying that $X$ is "close to" $Y$. So by counting the number of sets "close to" a given one, we can get a bound on the size of a family of sets whose pairwise symmetric differences are all "small."

Specifically, suppose $XsubseteqmathbbN$ (we might as well take $mathbbN$ to be our countable "base set"). Then there is a bijection between the sets close to $X$ and the finite sets of naturals - send a finite $F$ to the close-to-$X$ set $Xtriangle F$, and to invert this send a close-to-$X$ set $Y$ to the finite set $Xtriangle Y$ - and there are only countably many of the latter.

• An important point here is that $triangle$ allows cancellation: if $Xtriangle A=Xtriangle B$ then $A=B$. So the same finite difference can't occur more than once, and $Xtriangle F$ is the only set whose symmetric difference with $X$ is $F$ (and hence the map described above is surjective).

So any collection $mathcalF$ of subsets of $mathbbN$ (or any countable set) whose pairwise symmetric differences are finite must be countable: picking some $XinmathcalF$, there are only countably many subsets of $mathbbN$ whose symmetric difference with $X$ is finite.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Noah SchweberNoah Schweber

137k1010 gold badges163163 silver badges310310 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Am I right that you mean that any finite symmetric difference (there are only countably many of them) can be repeated only countably many times by any two distinct sets that creating a sub-family of sets which have this specific symmetric difference?
  $endgroup$
  – Viktor Zayakin
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorZayakin No - it can't be repeated at all! If $Xtriangle A =X triangle B$ then $A=B$. Each finite set corresponds to exactly one set close to $X$.
  $endgroup$
  – Noah Schweber
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorZayakin But that's not what I'm doing. I'm saying fix a single element $X$ of $mathcalF$ and think about things from $X$'s perspective. Everything in $mathcalF$ has to be close to $X$, but there are only countably many things close to $X$ in the first place.
  $endgroup$
  – Noah Schweber
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course, but it is possible that $mathitX triangle A = Y triangle B$ for some four sets. Sorry for using complicated wording (I'm not fluent enough). By the way, your last comment cleared it all for me, thanks a lot. This problem is done... Moving on!
  $endgroup$
  – Viktor Zayakin
  8 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

No, this is not possible.

Think about it this way: saying that $Xtriangle Y$ is "small" is saying that $X$ is "close to" $Y$. So by counting the number of sets "close to" a given one, we can get a bound on the size of a family of sets whose pairwise symmetric differences are all "small."

Specifically, suppose $XsubseteqmathbbN$ (we might as well take $mathbbN$ to be our countable "base set"). Then there is a bijection between the sets close to $X$ and the finite sets of naturals - send a finite $F$ to the close-to-$X$ set $Xtriangle F$, and to invert this send a close-to-$X$ set $Y$ to the finite set $Xtriangle Y$ - and there are only countably many of the latter.

• An important point here is that $triangle$ allows cancellation: if $Xtriangle A=Xtriangle B$ then $A=B$. So the same finite difference can't occur more than once, and $Xtriangle F$ is the only set whose symmetric difference with $X$ is $F$ (and hence the map described above is surjective).

So any collection $mathcalF$ of subsets of $mathbbN$ (or any countable set) whose pairwise symmetric differences are finite must be countable: picking some $XinmathcalF$, there are only countably many subsets of $mathbbN$ whose symmetric difference with $X$ is finite.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Noah SchweberNoah Schweber

137k1010 gold badges163163 silver badges310310 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Noah SchweberNoah Schweber

137k1010 gold badges163163 silver badges310310 bronze badges

answered 8 hours ago

Noah SchweberNoah Schweber

137k1010 gold badges163163 silver badges310310 bronze badges

answered 8 hours ago

Noah SchweberNoah Schweber

137k1010 gold badges163163 silver badges310310 bronze badges