Is $Bbb Q$ disconnected under general topology from $Bbb R$? The 2019 Stack Overflow Developer Survey Results Are InLet $A,B$ be nonempty subsets of a topological space $X$. Prove that $Acup B$ is disconnected if $(barAcap B)cup(AcapbarB)=emptyset$.Prove that $E$ is disconnected iff there exists two open disjoint sets $A$,$B$ in $X$$[0,1]times[0,1]$ stays connected after removal of an interior point$A cup B$ and $A cap B$ connected $implies A$ and $B$ are connectedConnected sets and separationsProve that the Sorgenfrey line is totally disconnectedA topological space $(X,Ƭ)$ is disconnected if it has at least one nonempty, proper subset that is both open and closed.Topology Proof Advice/GuidanceShow that the closure of a connected set is connectedProving $cup_jin J A_j$ is connected.
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Is $Bbb Q$ disconnected under general topology from $Bbb R$?
The 2019 Stack Overflow Developer Survey Results Are InLet $A,B$ be nonempty subsets of a topological space $X$. Prove that $Acup B$ is disconnected if $(barAcap B)cup(AcapbarB)=emptyset$.Prove that $E$ is disconnected iff there exists two open disjoint sets $A$,$B$ in $X$$[0,1]times[0,1]$ stays connected after removal of an interior point$A cup B$ and $A cap B$ connected $implies A$ and $B$ are connectedConnected sets and separationsProve that the Sorgenfrey line is totally disconnectedA topological space $(X,Ƭ)$ is disconnected if it has at least one nonempty, proper subset that is both open and closed.Topology Proof Advice/GuidanceShow that the closure of a connected set is connectedProving $cup_jin J A_j$ is connected.
$begingroup$
I just completed the proof of that fact that $Bbb Q$ is disconnected. My question is how to find a separation of $Bbb Q$.
Let $X$ be a topological space. We say $A$ and $B$ forms a separation of $X$ if $Acap barB$ and $barAcap B$ are empty and $Acup B=X$.
Here $Bbb Q$ denotes the set of rational numbers.
Can someone please help with hints.
Thanks.
general-topology
edited Mar 23 at 8:19
Andrews
1,2812423
asked Mar 23 at 3:54
StammeringMathematicianStammeringMathematician
2,7951324
$endgroup$
|
show 3 more comments
$begingroup$
I just completed the proof of that fact that $Bbb Q$ is disconnected. My question is how to find a separation of $Bbb Q$.
Let $X$ be a topological space. We say $A$ and $B$ forms a separation of $X$ if $Acap barB$ and $barAcap B$ are empty and $Acup B=X$.
Here $Bbb Q$ denotes the set of rational numbers.
Can someone please help with hints.
Thanks.
general-topology
edited Mar 23 at 8:19
Andrews
1,2812423
asked Mar 23 at 3:54
StammeringMathematicianStammeringMathematician
2,7951324
$endgroup$
2
$begingroup$
I think you are missing some properties... “two subsets which are disjoint from each other[‘s] closure” is surely not enough, or else $0$ and $1$ would have given you an answer. Perhaps start by having the correct definition of “separated”.
$endgroup$
– Arturo Magidin
Mar 23 at 3:55
$begingroup$
Please see the edit. @arturo
$endgroup$
– StammeringMathematician
Mar 23 at 3:58
$begingroup$
As a hint, the real numbers $mathbb R$ are connected and do not have the kind of "separation" you seek for $mathbb Q$. So the crux of the "separation" involves picking an irrational number (to split the rationals).
$endgroup$
– hardmath
Mar 23 at 4:00
2
$begingroup$
@hardmath I tried something like $(-infty,sqrt2)cup (sqrt2, infty)$
$endgroup$
– StammeringMathematician
Mar 23 at 4:03
1
$begingroup$
Yes, that is the classic example!
$endgroup$
– hardmath
Mar 23 at 4:05
|
show 3 more comments
$begingroup$
I just completed the proof of that fact that $Bbb Q$ is disconnected. My question is how to find a separation of $Bbb Q$.
Let $X$ be a topological space. We say $A$ and $B$ forms a separation of $X$ if $Acap barB$ and $barAcap B$ are empty and $Acup B=X$.
Here $Bbb Q$ denotes the set of rational numbers.
Can someone please help with hints.
Thanks.
general-topology
edited Mar 23 at 8:19
Andrews
1,2812423
asked Mar 23 at 3:54
StammeringMathematicianStammeringMathematician
2,7951324
$endgroup$
I just completed the proof of that fact that $Bbb Q$ is disconnected. My question is how to find a separation of $Bbb Q$.
Let $X$ be a topological space. We say $A$ and $B$ forms a separation of $X$ if $Acap barB$ and $barAcap B$ are empty and $Acup B=X$.
Here $Bbb Q$ denotes the set of rational numbers.
Can someone please help with hints.
Thanks.
general-topology
general-topology
edited Mar 23 at 8:19
Andrews
1,2812423
asked Mar 23 at 3:54
StammeringMathematicianStammeringMathematician
2,7951324
edited Mar 23 at 8:19
Andrews
1,2812423
asked Mar 23 at 3:54
StammeringMathematicianStammeringMathematician
2,7951324
edited Mar 23 at 8:19
Andrews
1,2812423
edited Mar 23 at 8:19
Andrews
1,2812423
edited Mar 23 at 8:19
Andrews
1,2812423
1,2812423
asked Mar 23 at 3:54
StammeringMathematicianStammeringMathematician
2,7951324
asked Mar 23 at 3:54
StammeringMathematicianStammeringMathematician
2,7951324
asked Mar 23 at 3:54
StammeringMathematicianStammeringMathematician
2,7951324
2,7951324
2
$begingroup$
I think you are missing some properties... “two subsets which are disjoint from each other[‘s] closure” is surely not enough, or else $0$ and $1$ would have given you an answer. Perhaps start by having the correct definition of “separated”.
$endgroup$
– Arturo Magidin
Mar 23 at 3:55
$begingroup$
Please see the edit. @arturo
$endgroup$
– StammeringMathematician
Mar 23 at 3:58
$begingroup$
As a hint, the real numbers $mathbb R$ are connected and do not have the kind of "separation" you seek for $mathbb Q$. So the crux of the "separation" involves picking an irrational number (to split the rationals).
$endgroup$
– hardmath
Mar 23 at 4:00
2
$begingroup$
@hardmath I tried something like $(-infty,sqrt2)cup (sqrt2, infty)$
$endgroup$
– StammeringMathematician
Mar 23 at 4:03
1
$begingroup$
Yes, that is the classic example!
$endgroup$
– hardmath
Mar 23 at 4:05
|
show 3 more comments
2
$begingroup$
I think you are missing some properties... “two subsets which are disjoint from each other[‘s] closure” is surely not enough, or else $0$ and $1$ would have given you an answer. Perhaps start by having the correct definition of “separated”.
$endgroup$
– Arturo Magidin
Mar 23 at 3:55
$begingroup$
Please see the edit. @arturo
$endgroup$
– StammeringMathematician
Mar 23 at 3:58
$begingroup$
As a hint, the real numbers $mathbb R$ are connected and do not have the kind of "separation" you seek for $mathbb Q$. So the crux of the "separation" involves picking an irrational number (to split the rationals).
$endgroup$
– hardmath
Mar 23 at 4:00
2
$begingroup$
@hardmath I tried something like $(-infty,sqrt2)cup (sqrt2, infty)$
$endgroup$
– StammeringMathematician
Mar 23 at 4:03
1
$begingroup$
Yes, that is the classic example!
$endgroup$
– hardmath
Mar 23 at 4:05
2
2
$begingroup$
I think you are missing some properties... “two subsets which are disjoint from each other[‘s] closure” is surely not enough, or else $0$ and $1$ would have given you an answer. Perhaps start by having the correct definition of “separated”.
$endgroup$
– Arturo Magidin
Mar 23 at 3:55
$begingroup$
I think you are missing some properties... “two subsets which are disjoint from each other[‘s] closure” is surely not enough, or else $0$ and $1$ would have given you an answer. Perhaps start by having the correct definition of “separated”.
$endgroup$
– Arturo Magidin
Mar 23 at 3:55
$begingroup$
Please see the edit. @arturo
$endgroup$
– StammeringMathematician
Mar 23 at 3:58
$begingroup$
Please see the edit. @arturo
$endgroup$
– StammeringMathematician
Mar 23 at 3:58
$begingroup$
As a hint, the real numbers $mathbb R$ are connected and do not have the kind of "separation" you seek for $mathbb Q$. So the crux of the "separation" involves picking an irrational number (to split the rationals).
$endgroup$
– hardmath
Mar 23 at 4:00
$begingroup$
As a hint, the real numbers $mathbb R$ are connected and do not have the kind of "separation" you seek for $mathbb Q$. So the crux of the "separation" involves picking an irrational number (to split the rationals).
$endgroup$
– hardmath
Mar 23 at 4:00
2
2
$begingroup$
@hardmath I tried something like $(-infty,sqrt2)cup (sqrt2, infty)$
$endgroup$
– StammeringMathematician
Mar 23 at 4:03
$begingroup$
@hardmath I tried something like $(-infty,sqrt2)cup (sqrt2, infty)$
$endgroup$
– StammeringMathematician
Mar 23 at 4:03
1
1
$begingroup$
Yes, that is the classic example!
$endgroup$
– hardmath
Mar 23 at 4:05
$begingroup$
Yes, that is the classic example!
$endgroup$
– hardmath
Mar 23 at 4:05
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$U= (-infty, sqrt2) cap mathbbQ$, is open in $mathbbQ$ as the intersection of an open subset of $mathbbR$ with $mathbbQ$ (definition of subspace topology) and similarly $V=(sqrt2, +infty) cap mathbbQ$ is also open in $mathbbQ$.
As $sqrt2 notin mathbbQ$ we have $mathbbQ = U cup V$ and $overlineU = U$ (closure in $mathbbQ$) and $overlineV=V$ and clearly $U cap V=emptyset$, so $U,V$ forms a separation of $mathbbQ$.
answered Mar 23 at 5:50
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$U= (-infty, sqrt2) cap mathbbQ$, is open in $mathbbQ$ as the intersection of an open subset of $mathbbR$ with $mathbbQ$ (definition of subspace topology) and similarly $V=(sqrt2, +infty) cap mathbbQ$ is also open in $mathbbQ$.
As $sqrt2 notin mathbbQ$ we have $mathbbQ = U cup V$ and $overlineU = U$ (closure in $mathbbQ$) and $overlineV=V$ and clearly $U cap V=emptyset$, so $U,V$ forms a separation of $mathbbQ$.
answered Mar 23 at 5:50
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
$U= (-infty, sqrt2) cap mathbbQ$, is open in $mathbbQ$ as the intersection of an open subset of $mathbbR$ with $mathbbQ$ (definition of subspace topology) and similarly $V=(sqrt2, +infty) cap mathbbQ$ is also open in $mathbbQ$.
As $sqrt2 notin mathbbQ$ we have $mathbbQ = U cup V$ and $overlineU = U$ (closure in $mathbbQ$) and $overlineV=V$ and clearly $U cap V=emptyset$, so $U,V$ forms a separation of $mathbbQ$.
answered Mar 23 at 5:50
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
$U= (-infty, sqrt2) cap mathbbQ$, is open in $mathbbQ$ as the intersection of an open subset of $mathbbR$ with $mathbbQ$ (definition of subspace topology) and similarly $V=(sqrt2, +infty) cap mathbbQ$ is also open in $mathbbQ$.
As $sqrt2 notin mathbbQ$ we have $mathbbQ = U cup V$ and $overlineU = U$ (closure in $mathbbQ$) and $overlineV=V$ and clearly $U cap V=emptyset$, so $U,V$ forms a separation of $mathbbQ$.
answered Mar 23 at 5:50
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
$U= (-infty, sqrt2) cap mathbbQ$, is open in $mathbbQ$ as the intersection of an open subset of $mathbbR$ with $mathbbQ$ (definition of subspace topology) and similarly $V=(sqrt2, +infty) cap mathbbQ$ is also open in $mathbbQ$.
As $sqrt2 notin mathbbQ$ we have $mathbbQ = U cup V$ and $overlineU = U$ (closure in $mathbbQ$) and $overlineV=V$ and clearly $U cap V=emptyset$, so $U,V$ forms a separation of $mathbbQ$.
answered Mar 23 at 5:50
Henno BrandsmaHenno Brandsma
116k349127
answered Mar 23 at 5:50
Henno BrandsmaHenno Brandsma
116k349127
answered Mar 23 at 5:50
Henno BrandsmaHenno Brandsma
116k349127
answered Mar 23 at 5:50
Henno BrandsmaHenno Brandsma
116k349127
116k349127
add a comment |
add a comment |
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$begingroup$
I think you are missing some properties... “two subsets which are disjoint from each other[‘s] closure” is surely not enough, or else $0$ and $1$ would have given you an answer. Perhaps start by having the correct definition of “separated”.
$endgroup$
– Arturo Magidin
Mar 23 at 3:55
$begingroup$
Please see the edit. @arturo
$endgroup$
– StammeringMathematician
Mar 23 at 3:58
$begingroup$
As a hint, the real numbers $mathbb R$ are connected and do not have the kind of "separation" you seek for $mathbb Q$. So the crux of the "separation" involves picking an irrational number (to split the rationals).
$endgroup$
– hardmath
Mar 23 at 4:00
2
$begingroup$
@hardmath I tried something like $(-infty,sqrt2)cup (sqrt2, infty)$
$endgroup$
– StammeringMathematician
Mar 23 at 4:03
1
$begingroup$
Yes, that is the classic example!
$endgroup$
– hardmath
Mar 23 at 4:05