Show $exists X,Y,Z subset S$ disjoint and $|X|=|Y|=|Z|=c$ The 2019 Stack Overflow Developer Survey Results Are Inconvert union of sets into union of disjoint setsReducing countable collections of countably infinite sets to pairwise disjoint setsMutually disjoint implying complements in set theoryCardinality of the union of disjoint sets, each of which have a cardinality of realsSubsets of a Cartesian product are disjoint iff there exist projections that are disjointProve existence a maximal pairwise disjoint subfamily $mathcalS subseteq mathcalR$Show that a finite subset of the reals exists whose sum is greater than some real numberIn set theory, to prove $A=B$, is it always necessary to prove that $Asubset B$ and $Bsubset A$How can we show a set S and a subset of S have the same cardinality? Specifically: (below)Can $[0,1]times[0,1]$ be partitioned into uncountably many disjoint non-degenerate intervals?
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Show $exists X,Y,Z subset S$ disjoint and $|X|=|Y|=|Z|=c$
The 2019 Stack Overflow Developer Survey Results Are Inconvert union of sets into union of disjoint setsReducing countable collections of countably infinite sets to pairwise disjoint setsMutually disjoint implying complements in set theoryCardinality of the union of disjoint sets, each of which have a cardinality of realsSubsets of a Cartesian product are disjoint iff there exist projections that are disjointProve existence a maximal pairwise disjoint subfamily $mathcalS subseteq mathcalR$Show that a finite subset of the reals exists whose sum is greater than some real numberIn set theory, to prove $A=B$, is it always necessary to prove that $Asubset B$ and $Bsubset A$How can we show a set S and a subset of S have the same cardinality? Specifically: (below)Can $[0,1]times[0,1]$ be partitioned into uncountably many disjoint non-degenerate intervals?
$begingroup$
Given $|S| = c$ and $c$ here represents the continuum.
Given $x,y,z in S$ distinct. Want to show $exists X,Y,Z subset S$ such that $|X|=|Y|=|Z|=c$ and $x in X, y in Y, z in Z$.
I tried defining $f: S to [0,1]$ and given $x,y,zin S$ $f(x),f(y),f(z)in[0,1].$
I can always find a subinterval of $[0,1]$ such that they are disjoint and $x,y,z$ is in each subinterval.
Would this be a correct proof idea? Or am I approaching this wrong... Thanks.
elementary-set-theory
edited Mar 23 at 13:16
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 3:45
javacoderjavacoder
848
$endgroup$
|
show 8 more comments
$begingroup$
Given $|S| = c$ and $c$ here represents the continuum.
Given $x,y,z in S$ distinct. Want to show $exists X,Y,Z subset S$ such that $|X|=|Y|=|Z|=c$ and $x in X, y in Y, z in Z$.
I tried defining $f: S to [0,1]$ and given $x,y,zin S$ $f(x),f(y),f(z)in[0,1].$
I can always find a subinterval of $[0,1]$ such that they are disjoint and $x,y,z$ is in each subinterval.
Would this be a correct proof idea? Or am I approaching this wrong... Thanks.
elementary-set-theory
edited Mar 23 at 13:16
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 3:45
javacoderjavacoder
848
$endgroup$
1
$begingroup$
I find it hard to believe that you would be able to define a function $f$. I would find it more credible if you somehow were able to guarantee the existence of such an $f$, and presumably imbue $f$ with some specific properties (one-to-one? onto? both? neither?) from which you could deduce the existence of the desired $X$, $Y$, and $Z$. Also, $x$,$y$, and $z$ do not “live” in $[0,1]$, so you can’t find subintervals with them...
$endgroup$
– Arturo Magidin
Mar 23 at 3:54
$begingroup$
Okay I know there exists a bijection, sorry for the poor wording by saying define. What do I do next?
$endgroup$
– javacoder
Mar 23 at 4:25
$begingroup$
Your idea is fine, but you need to flesh it out and complete it. And be careful; again, you claim something about $x$, $y$, and $z$ as if they were elements of $[0,1]$. They aren’t.
$endgroup$
– Arturo Magidin
Mar 23 at 4:33
$begingroup$
I meant $f(x), f(y), f(z) in [0,1]$ I can find disjoint subsets of $[0,1]$ for each $f(x),f(y),f(z)$ but how do I make it for $S$? Maybe because $f$ is bijective so I can do something but not sure what.
$endgroup$
– javacoder
Mar 23 at 4:38
$begingroup$
It’s not enought that they be subsets; you need them to have a certain cardinality. As to “not sure what”... what is the defining characteristic of a bijecive function? It has an inverse.
$endgroup$
– Arturo Magidin
Mar 23 at 4:39
|
show 8 more comments
$begingroup$
Given $|S| = c$ and $c$ here represents the continuum.
Given $x,y,z in S$ distinct. Want to show $exists X,Y,Z subset S$ such that $|X|=|Y|=|Z|=c$ and $x in X, y in Y, z in Z$.
I tried defining $f: S to [0,1]$ and given $x,y,zin S$ $f(x),f(y),f(z)in[0,1].$
I can always find a subinterval of $[0,1]$ such that they are disjoint and $x,y,z$ is in each subinterval.
Would this be a correct proof idea? Or am I approaching this wrong... Thanks.
elementary-set-theory
edited Mar 23 at 13:16
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 3:45
javacoderjavacoder
848
$endgroup$
Given $|S| = c$ and $c$ here represents the continuum.
Given $x,y,z in S$ distinct. Want to show $exists X,Y,Z subset S$ such that $|X|=|Y|=|Z|=c$ and $x in X, y in Y, z in Z$.
I tried defining $f: S to [0,1]$ and given $x,y,zin S$ $f(x),f(y),f(z)in[0,1].$
I can always find a subinterval of $[0,1]$ such that they are disjoint and $x,y,z$ is in each subinterval.
Would this be a correct proof idea? Or am I approaching this wrong... Thanks.
elementary-set-theory
elementary-set-theory
edited Mar 23 at 13:16
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 3:45
javacoderjavacoder
848
edited Mar 23 at 13:16
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 3:45
javacoderjavacoder
848
edited Mar 23 at 13:16
Andrés E. Caicedo
65.9k8160252
edited Mar 23 at 13:16
Andrés E. Caicedo
65.9k8160252
edited Mar 23 at 13:16
Andrés E. Caicedo
65.9k8160252
65.9k8160252
asked Mar 23 at 3:45
javacoderjavacoder
848
asked Mar 23 at 3:45
javacoderjavacoder
848
asked Mar 23 at 3:45
javacoderjavacoder
848
848
1
$begingroup$
I find it hard to believe that you would be able to define a function $f$. I would find it more credible if you somehow were able to guarantee the existence of such an $f$, and presumably imbue $f$ with some specific properties (one-to-one? onto? both? neither?) from which you could deduce the existence of the desired $X$, $Y$, and $Z$. Also, $x$,$y$, and $z$ do not “live” in $[0,1]$, so you can’t find subintervals with them...
$endgroup$
– Arturo Magidin
Mar 23 at 3:54
$begingroup$
Okay I know there exists a bijection, sorry for the poor wording by saying define. What do I do next?
$endgroup$
– javacoder
Mar 23 at 4:25
$begingroup$
Your idea is fine, but you need to flesh it out and complete it. And be careful; again, you claim something about $x$, $y$, and $z$ as if they were elements of $[0,1]$. They aren’t.
$endgroup$
– Arturo Magidin
Mar 23 at 4:33
$begingroup$
I meant $f(x), f(y), f(z) in [0,1]$ I can find disjoint subsets of $[0,1]$ for each $f(x),f(y),f(z)$ but how do I make it for $S$? Maybe because $f$ is bijective so I can do something but not sure what.
$endgroup$
– javacoder
Mar 23 at 4:38
$begingroup$
It’s not enought that they be subsets; you need them to have a certain cardinality. As to “not sure what”... what is the defining characteristic of a bijecive function? It has an inverse.
$endgroup$
– Arturo Magidin
Mar 23 at 4:39
|
show 8 more comments
1
$begingroup$
I find it hard to believe that you would be able to define a function $f$. I would find it more credible if you somehow were able to guarantee the existence of such an $f$, and presumably imbue $f$ with some specific properties (one-to-one? onto? both? neither?) from which you could deduce the existence of the desired $X$, $Y$, and $Z$. Also, $x$,$y$, and $z$ do not “live” in $[0,1]$, so you can’t find subintervals with them...
$endgroup$
– Arturo Magidin
Mar 23 at 3:54
$begingroup$
Okay I know there exists a bijection, sorry for the poor wording by saying define. What do I do next?
$endgroup$
– javacoder
Mar 23 at 4:25
$begingroup$
Your idea is fine, but you need to flesh it out and complete it. And be careful; again, you claim something about $x$, $y$, and $z$ as if they were elements of $[0,1]$. They aren’t.
$endgroup$
– Arturo Magidin
Mar 23 at 4:33
$begingroup$
I meant $f(x), f(y), f(z) in [0,1]$ I can find disjoint subsets of $[0,1]$ for each $f(x),f(y),f(z)$ but how do I make it for $S$? Maybe because $f$ is bijective so I can do something but not sure what.
$endgroup$
– javacoder
Mar 23 at 4:38
$begingroup$
It’s not enought that they be subsets; you need them to have a certain cardinality. As to “not sure what”... what is the defining characteristic of a bijecive function? It has an inverse.
$endgroup$
– Arturo Magidin
Mar 23 at 4:39
1
1
$begingroup$
I find it hard to believe that you would be able to define a function $f$. I would find it more credible if you somehow were able to guarantee the existence of such an $f$, and presumably imbue $f$ with some specific properties (one-to-one? onto? both? neither?) from which you could deduce the existence of the desired $X$, $Y$, and $Z$. Also, $x$,$y$, and $z$ do not “live” in $[0,1]$, so you can’t find subintervals with them...
$endgroup$
– Arturo Magidin
Mar 23 at 3:54
$begingroup$
I find it hard to believe that you would be able to define a function $f$. I would find it more credible if you somehow were able to guarantee the existence of such an $f$, and presumably imbue $f$ with some specific properties (one-to-one? onto? both? neither?) from which you could deduce the existence of the desired $X$, $Y$, and $Z$. Also, $x$,$y$, and $z$ do not “live” in $[0,1]$, so you can’t find subintervals with them...
$endgroup$
– Arturo Magidin
Mar 23 at 3:54
$begingroup$
Okay I know there exists a bijection, sorry for the poor wording by saying define. What do I do next?
$endgroup$
– javacoder
Mar 23 at 4:25
$begingroup$
Okay I know there exists a bijection, sorry for the poor wording by saying define. What do I do next?
$endgroup$
– javacoder
Mar 23 at 4:25
$begingroup$
Your idea is fine, but you need to flesh it out and complete it. And be careful; again, you claim something about $x$, $y$, and $z$ as if they were elements of $[0,1]$. They aren’t.
$endgroup$
– Arturo Magidin
Mar 23 at 4:33
$begingroup$
Your idea is fine, but you need to flesh it out and complete it. And be careful; again, you claim something about $x$, $y$, and $z$ as if they were elements of $[0,1]$. They aren’t.
$endgroup$
– Arturo Magidin
Mar 23 at 4:33
$begingroup$
I meant $f(x), f(y), f(z) in [0,1]$ I can find disjoint subsets of $[0,1]$ for each $f(x),f(y),f(z)$ but how do I make it for $S$? Maybe because $f$ is bijective so I can do something but not sure what.
$endgroup$
– javacoder
Mar 23 at 4:38
$begingroup$
I meant $f(x), f(y), f(z) in [0,1]$ I can find disjoint subsets of $[0,1]$ for each $f(x),f(y),f(z)$ but how do I make it for $S$? Maybe because $f$ is bijective so I can do something but not sure what.
$endgroup$
– javacoder
Mar 23 at 4:38
$begingroup$
It’s not enought that they be subsets; you need them to have a certain cardinality. As to “not sure what”... what is the defining characteristic of a bijecive function? It has an inverse.
$endgroup$
– Arturo Magidin
Mar 23 at 4:39
$begingroup$
It’s not enought that they be subsets; you need them to have a certain cardinality. As to “not sure what”... what is the defining characteristic of a bijecive function? It has an inverse.
$endgroup$
– Arturo Magidin
Mar 23 at 4:39
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
More precisely there exists a bijection $f:Sto Bbb R $ and $r= frac 12min in Bbb R^+.$
So let $X=f^-1(-r+f(x),r+f(x))$ and $Y=f^-1(-r+f(y),r+f(y))$ and $Z=f^-1(-r+f(z),r+f(z)).$
answered Mar 23 at 14:46
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
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$begingroup$
More precisely there exists a bijection $f:Sto Bbb R $ and $r= frac 12min in Bbb R^+.$
So let $X=f^-1(-r+f(x),r+f(x))$ and $Y=f^-1(-r+f(y),r+f(y))$ and $Z=f^-1(-r+f(z),r+f(z)).$
answered Mar 23 at 14:46
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
More precisely there exists a bijection $f:Sto Bbb R $ and $r= frac 12min in Bbb R^+.$
So let $X=f^-1(-r+f(x),r+f(x))$ and $Y=f^-1(-r+f(y),r+f(y))$ and $Z=f^-1(-r+f(z),r+f(z)).$
answered Mar 23 at 14:46
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
More precisely there exists a bijection $f:Sto Bbb R $ and $r= frac 12min in Bbb R^+.$
So let $X=f^-1(-r+f(x),r+f(x))$ and $Y=f^-1(-r+f(y),r+f(y))$ and $Z=f^-1(-r+f(z),r+f(z)).$
answered Mar 23 at 14:46
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
More precisely there exists a bijection $f:Sto Bbb R $ and $r= frac 12min in Bbb R^+.$
So let $X=f^-1(-r+f(x),r+f(x))$ and $Y=f^-1(-r+f(y),r+f(y))$ and $Z=f^-1(-r+f(z),r+f(z)).$
answered Mar 23 at 14:46
DanielWainfleetDanielWainfleet
35.8k31648
answered Mar 23 at 14:46
DanielWainfleetDanielWainfleet
35.8k31648
answered Mar 23 at 14:46
DanielWainfleetDanielWainfleet
35.8k31648
answered Mar 23 at 14:46
DanielWainfleetDanielWainfleet
35.8k31648
35.8k31648
add a comment |
add a comment |
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1
$begingroup$
I find it hard to believe that you would be able to define a function $f$. I would find it more credible if you somehow were able to guarantee the existence of such an $f$, and presumably imbue $f$ with some specific properties (one-to-one? onto? both? neither?) from which you could deduce the existence of the desired $X$, $Y$, and $Z$. Also, $x$,$y$, and $z$ do not “live” in $[0,1]$, so you can’t find subintervals with them...
$endgroup$
– Arturo Magidin
Mar 23 at 3:54
$begingroup$
Okay I know there exists a bijection, sorry for the poor wording by saying define. What do I do next?
$endgroup$
– javacoder
Mar 23 at 4:25
$begingroup$
Your idea is fine, but you need to flesh it out and complete it. And be careful; again, you claim something about $x$, $y$, and $z$ as if they were elements of $[0,1]$. They aren’t.
$endgroup$
– Arturo Magidin
Mar 23 at 4:33
$begingroup$
I meant $f(x), f(y), f(z) in [0,1]$ I can find disjoint subsets of $[0,1]$ for each $f(x),f(y),f(z)$ but how do I make it for $S$? Maybe because $f$ is bijective so I can do something but not sure what.
$endgroup$
– javacoder
Mar 23 at 4:38
$begingroup$
It’s not enought that they be subsets; you need them to have a certain cardinality. As to “not sure what”... what is the defining characteristic of a bijecive function? It has an inverse.
$endgroup$
– Arturo Magidin
Mar 23 at 4:39