A sequence $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Separability of a product metric spaceProve equivalent metric spacesConcept of Cauchy sequenceProve that all three metrics induces the same topology on $X_1times X_2$If $(X,d_1)$ and $(X,d_2)$ two connected metric spaces if only if $Xtimes Y$ is connected metric spaceCompletion of two metric spacesProve that an open ball in $(X,d)$ is the product of open balls from $X_1, X_2, ldots, X_n$.If $d_1(x,y) leq d_2(x,y)$ and all $d_2$-closed balls in $M$ are also $d_1$-closed, then $(M,d_2)$ is complete.$x_n rightarrow x$ iff the modified sequence is CauchyLet $tau$ denote the product topology on $X=X_1 times X_2$. Then $(X,tau)$ is a separable space.
Sort list of array linked objects by keys and values
how can a perfect fourth interval be considered either consonant or dissonant?
Simulating Exploding Dice
How many people can fit inside Mordenkainen's Magnificent Mansion?
Does the AirPods case need to be around while listening via an iOS Device?
ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?
Road tyres vs "Street" tyres for charity ride on MTB Tandem
Is a pteranodon too powerful as a beast companion for a beast master?
Relations between two reciprocal partial derivatives?
He got a vote 80% that of Emmanuel Macron’s
How can I protect witches in combat who wear limited clothing?
Why not take a picture of a closer black hole?
Match Roman Numerals
Derivation tree not rendering
Can a 1st-level character have an ability score above 18?
Keeping a retro style to sci-fi spaceships?
Can withdrawing asylum be illegal?
How do I add random spotting to the same face in cycles?
Did the new image of black hole confirm the general theory of relativity?
Finding the path in a graph from A to B then back to A with a minimum of shared edges
Empty set is subset of every set? If yes, why that...
Why does the Event Horizon Telescope (EHT) not include telescopes from Africa, Asia or Australia?
Change bounding box of math glyphs in LuaTeX
Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?
A sequence $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Separability of a product metric spaceProve equivalent metric spacesConcept of Cauchy sequenceProve that all three metrics induces the same topology on $X_1times X_2$If $(X,d_1)$ and $(X,d_2)$ two connected metric spaces if only if $Xtimes Y$ is connected metric spaceCompletion of two metric spacesProve that an open ball in $(X,d)$ is the product of open balls from $X_1, X_2, ldots, X_n$.If $d_1(x,y) leq d_2(x,y)$ and all $d_2$-closed balls in $M$ are also $d_1$-closed, then $(M,d_2)$ is complete.$x_n rightarrow x$ iff the modified sequence is CauchyLet $tau$ denote the product topology on $X=X_1 times X_2$. Then $(X,tau)$ is a separable space.
$begingroup$
Let $X=(mathbbR^k,d)$ denote the standard n-Euclidean space.
Now I read a theorem which states that A sequence $x_n=(x_1,n,x_2,n,x_3,n...x_k,n)$ in $X$ is Cauchy if and only if $x_j,n$ is Cauchy in $R$
But can we generalize this result?
Question : Does this result hold for general metric spaces, and if not, what conditions will guarantee that it does?
I mean take $X_1, d_1$ and $X_2,d_2$ be any two metric space. Let $X=X_1times X_2$ and let $d$ be any metric on $X$( not necessarily a product metric )
then can we say that a $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?
As far as I know, being Cauchy is a property of metric, so the result may not hold in general.
Can some one please clarify? Are there some counterexamples.
Thanks a lot.
real-analysis general-topology
$endgroup$
add a comment |
$begingroup$
Let $X=(mathbbR^k,d)$ denote the standard n-Euclidean space.
Now I read a theorem which states that A sequence $x_n=(x_1,n,x_2,n,x_3,n...x_k,n)$ in $X$ is Cauchy if and only if $x_j,n$ is Cauchy in $R$
But can we generalize this result?
Question : Does this result hold for general metric spaces, and if not, what conditions will guarantee that it does?
I mean take $X_1, d_1$ and $X_2,d_2$ be any two metric space. Let $X=X_1times X_2$ and let $d$ be any metric on $X$( not necessarily a product metric )
then can we say that a $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?
As far as I know, being Cauchy is a property of metric, so the result may not hold in general.
Can some one please clarify? Are there some counterexamples.
Thanks a lot.
real-analysis general-topology
$endgroup$
$begingroup$
@KajHansen Thanks.
$endgroup$
– StammeringMathematician
Apr 8 at 12:22
add a comment |
$begingroup$
Let $X=(mathbbR^k,d)$ denote the standard n-Euclidean space.
Now I read a theorem which states that A sequence $x_n=(x_1,n,x_2,n,x_3,n...x_k,n)$ in $X$ is Cauchy if and only if $x_j,n$ is Cauchy in $R$
But can we generalize this result?
Question : Does this result hold for general metric spaces, and if not, what conditions will guarantee that it does?
I mean take $X_1, d_1$ and $X_2,d_2$ be any two metric space. Let $X=X_1times X_2$ and let $d$ be any metric on $X$( not necessarily a product metric )
then can we say that a $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?
As far as I know, being Cauchy is a property of metric, so the result may not hold in general.
Can some one please clarify? Are there some counterexamples.
Thanks a lot.
real-analysis general-topology
$endgroup$
Let $X=(mathbbR^k,d)$ denote the standard n-Euclidean space.
Now I read a theorem which states that A sequence $x_n=(x_1,n,x_2,n,x_3,n...x_k,n)$ in $X$ is Cauchy if and only if $x_j,n$ is Cauchy in $R$
But can we generalize this result?
Question : Does this result hold for general metric spaces, and if not, what conditions will guarantee that it does?
I mean take $X_1, d_1$ and $X_2,d_2$ be any two metric space. Let $X=X_1times X_2$ and let $d$ be any metric on $X$( not necessarily a product metric )
then can we say that a $(x_n,x_m)$ is Cauchy in $X=X_1times X_2$ if and only if $x_n$ is Cauchy in $X_1$ and $x_m$ is Cauchy in $X_2$?
As far as I know, being Cauchy is a property of metric, so the result may not hold in general.
Can some one please clarify? Are there some counterexamples.
Thanks a lot.
real-analysis general-topology
real-analysis general-topology
edited Apr 8 at 12:21
Kaj Hansen
27.8k43980
27.8k43980
asked Apr 8 at 11:20
StammeringMathematicianStammeringMathematician
2,7951324
2,7951324
$begingroup$
@KajHansen Thanks.
$endgroup$
– StammeringMathematician
Apr 8 at 12:22
add a comment |
$begingroup$
@KajHansen Thanks.
$endgroup$
– StammeringMathematician
Apr 8 at 12:22
$begingroup$
@KajHansen Thanks.
$endgroup$
– StammeringMathematician
Apr 8 at 12:22
$begingroup$
@KajHansen Thanks.
$endgroup$
– StammeringMathematician
Apr 8 at 12:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.
$endgroup$
$begingroup$
Does the backward implication hold, under some conditions ?
$endgroup$
– spkakkar
Apr 8 at 11:30
$begingroup$
What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
$endgroup$
– StammeringMathematician
Apr 8 at 11:33
$begingroup$
Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
$endgroup$
– Mark
Apr 8 at 11:45
add a comment |
$begingroup$
Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...
$endgroup$
1
$begingroup$
Thanks. I got it. Under what conditions does the result hold true? Thanks
$endgroup$
– StammeringMathematician
Apr 8 at 11:34
add a comment |
$begingroup$
When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
$d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
$$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.
$endgroup$
add a comment |
Your Answer
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%2f3179494%2fa-sequence-x-n-x-m-is-cauchy-in-x-x-1-times-x-2-if-and-only-if-x-n-is-ca%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.
$endgroup$
$begingroup$
Does the backward implication hold, under some conditions ?
$endgroup$
– spkakkar
Apr 8 at 11:30
$begingroup$
What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
$endgroup$
– StammeringMathematician
Apr 8 at 11:33
$begingroup$
Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
$endgroup$
– Mark
Apr 8 at 11:45
add a comment |
$begingroup$
If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.
$endgroup$
$begingroup$
Does the backward implication hold, under some conditions ?
$endgroup$
– spkakkar
Apr 8 at 11:30
$begingroup$
What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
$endgroup$
– StammeringMathematician
Apr 8 at 11:33
$begingroup$
Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
$endgroup$
– Mark
Apr 8 at 11:45
add a comment |
$begingroup$
If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.
$endgroup$
If you let $d$ be any metric on $X$ then of course the answer is no. Let's say we take $mathbbR$ with the usual metric and $mathbbR^2$ with the discrete metric. (i.e $d(x,y)=delta_xy)$.Then the sequence $(frac1n,frac1n)$ is not Cauchy in $mathbbR^2$ but in both coordinates you have Cauchy sequences.
answered Apr 8 at 11:28
MarkMark
10.5k1622
10.5k1622
$begingroup$
Does the backward implication hold, under some conditions ?
$endgroup$
– spkakkar
Apr 8 at 11:30
$begingroup$
What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
$endgroup$
– StammeringMathematician
Apr 8 at 11:33
$begingroup$
Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
$endgroup$
– Mark
Apr 8 at 11:45
add a comment |
$begingroup$
Does the backward implication hold, under some conditions ?
$endgroup$
– spkakkar
Apr 8 at 11:30
$begingroup$
What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
$endgroup$
– StammeringMathematician
Apr 8 at 11:33
$begingroup$
Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
$endgroup$
– Mark
Apr 8 at 11:45
$begingroup$
Does the backward implication hold, under some conditions ?
$endgroup$
– spkakkar
Apr 8 at 11:30
$begingroup$
Does the backward implication hold, under some conditions ?
$endgroup$
– spkakkar
Apr 8 at 11:30
$begingroup$
What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
$endgroup$
– StammeringMathematician
Apr 8 at 11:33
$begingroup$
What happens, if we take $d$ to be the product metric. I mean I am curious to understand why you have bold the term "any". It would be too kind of you if you can throw some light on it.
$endgroup$
– StammeringMathematician
Apr 8 at 11:33
$begingroup$
Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
$endgroup$
– Mark
Apr 8 at 11:45
$begingroup$
Yes, sometimes it does hold. If you take take a $p$-product metric then it works well. (the proof is just like in the standard euclidean space)
$endgroup$
– Mark
Apr 8 at 11:45
add a comment |
$begingroup$
Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...
$endgroup$
1
$begingroup$
Thanks. I got it. Under what conditions does the result hold true? Thanks
$endgroup$
– StammeringMathematician
Apr 8 at 11:34
add a comment |
$begingroup$
Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...
$endgroup$
1
$begingroup$
Thanks. I got it. Under what conditions does the result hold true? Thanks
$endgroup$
– StammeringMathematician
Apr 8 at 11:34
add a comment |
$begingroup$
Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...
$endgroup$
Taking the discrete metric on $X_1times X_2$ will probably be enough to find a counterexample...
answered Apr 8 at 11:28
5xum5xum
92.6k394162
92.6k394162
1
$begingroup$
Thanks. I got it. Under what conditions does the result hold true? Thanks
$endgroup$
– StammeringMathematician
Apr 8 at 11:34
add a comment |
1
$begingroup$
Thanks. I got it. Under what conditions does the result hold true? Thanks
$endgroup$
– StammeringMathematician
Apr 8 at 11:34
1
1
$begingroup$
Thanks. I got it. Under what conditions does the result hold true? Thanks
$endgroup$
– StammeringMathematician
Apr 8 at 11:34
$begingroup$
Thanks. I got it. Under what conditions does the result hold true? Thanks
$endgroup$
– StammeringMathematician
Apr 8 at 11:34
add a comment |
$begingroup$
When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
$d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
$$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.
$endgroup$
add a comment |
$begingroup$
When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
$d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
$$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.
$endgroup$
add a comment |
$begingroup$
When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
$d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
$$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.
$endgroup$
When considering a product of two metric spaces, say $(X_1,d_1)$ and $(X_2,d_2)$, you want to consider metrics on the set $X_1times X_2$ that are naturally related to the individual metrics $d_1,d_2$. For example, you may consider
$d((x,y),(xi,eta)):=d_1(x,xi)+d_2(y,eta)$. More generally, if you have a countable collection of metric spaces $(X_i,d_i)_i=1^infty$ you may define a metric on the product $prod_i=1^inftyX_i$ by
$$d((x_i),(y_i))=sum_i=1^inftyfracd_i(x_i,y_i)2^i$$
For such metrics on the product, which use the individual metrics in some "natural" way, it is the case that $(x_i,n)_n=1^infty$ is a Cauchy sequence in the product space if and only if each sequence $x_i,n$ is a Cauchy sequence in $X_i$.
answered Apr 8 at 12:33
uniquesolutionuniquesolution
9,5471823
9,5471823
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%2f3179494%2fa-sequence-x-n-x-m-is-cauchy-in-x-x-1-times-x-2-if-and-only-if-x-n-is-ca%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
$begingroup$
@KajHansen Thanks.
$endgroup$
– StammeringMathematician
Apr 8 at 12:22