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.
How to change the limits of integration
Why is Grand Jury testimony secret?
Limit the amount of RAM Mathematica may access?
Spanish for "widget"
Patience, young "Padovan"
What is the use of option -o in the useradd command?
Understanding the implication of what "well-defined" means for the operation in quotient group
Pristine Bit Checking
Monty Hall variation
How was Skylab's orbit inclination chosen?
aging parents with no investments
Why is the maximum length of OpenWrt’s root password 8 characters?
Could JWST stay at L2 "forever"?
Does it makes sense to buy a new cycle to learn riding?
If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?
What tool would a Roman-age civilization have to grind silver and other metals into dust?
Is bread bad for ducks?
Realistic Alternatives to Dust: What Else Could Feed a Plankton Bloom?
Landlord wants to switch my lease to a "Land contract" to "get back at the city"
Springs with some finite mass
Are there any other methods to apply to solving simultaneous equations?
Extreme, unacceptable situation and I can't attend work tomorrow morning
How to answer pointed "are you quitting" questioning when I don't want them to suspect
How to reverse every other sublist of a list?
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 |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158932%2fis-bbb-q-disconnected-under-general-topology-from-bbb-r%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 |
$begingroup$
$U= (-infty, sqrt2) cap mathbbQ$, 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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158932%2fis-bbb-q-disconnected-under-general-topology-from-bbb-r%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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