Prove that $U=x in Xmid d(x,A)<d(x,B)$ is open when $A$ and $B$ are disjoint
Clash Royale CLAN TAG#URR8PPP
$begingroup$
In a metric space $(X,d)$, I have the set $$U=x in Xmid d(x,A)<d(x,B)$$
where $A$ and $B$ are disjoint subsets. I need to show that $U$ is open in $(X,d)$. I tried taking the radius of an open ball centre $xin U$ to be less than $d(x,C)$ where $C=xin Xmid d(x,A)=d(x,C)$ but I could not get anywhere. Is this the right strategy? I would really appreciate help, thank you!
metric-spaces
edited Mar 14 at 19:47
Asaf Karagila♦
308k33441775
asked Mar 14 at 13:34
Sean ThrasherSean Thrasher
444
$endgroup$
add a comment |
$begingroup$
In a metric space $(X,d)$, I have the set $$U=x in Xmid d(x,A)<d(x,B)$$
where $A$ and $B$ are disjoint subsets. I need to show that $U$ is open in $(X,d)$. I tried taking the radius of an open ball centre $xin U$ to be less than $d(x,C)$ where $C=xin Xmid d(x,A)=d(x,C)$ but I could not get anywhere. Is this the right strategy? I would really appreciate help, thank you!
metric-spaces
edited Mar 14 at 19:47
Asaf Karagila♦
308k33441775
asked Mar 14 at 13:34
Sean ThrasherSean Thrasher
444
$endgroup$
add a comment |
$begingroup$
In a metric space $(X,d)$, I have the set $$U=x in Xmid d(x,A)<d(x,B)$$
where $A$ and $B$ are disjoint subsets. I need to show that $U$ is open in $(X,d)$. I tried taking the radius of an open ball centre $xin U$ to be less than $d(x,C)$ where $C=xin Xmid d(x,A)=d(x,C)$ but I could not get anywhere. Is this the right strategy? I would really appreciate help, thank you!
metric-spaces
edited Mar 14 at 19:47
Asaf Karagila♦
308k33441775
asked Mar 14 at 13:34
Sean ThrasherSean Thrasher
444
$endgroup$
In a metric space $(X,d)$, I have the set $$U=x in Xmid d(x,A)<d(x,B)$$
where $A$ and $B$ are disjoint subsets. I need to show that $U$ is open in $(X,d)$. I tried taking the radius of an open ball centre $xin U$ to be less than $d(x,C)$ where $C=xin Xmid d(x,A)=d(x,C)$ but I could not get anywhere. Is this the right strategy? I would really appreciate help, thank you!
metric-spaces
metric-spaces
edited Mar 14 at 19:47
Asaf Karagila♦
308k33441775
asked Mar 14 at 13:34
Sean ThrasherSean Thrasher
444
edited Mar 14 at 19:47
Asaf Karagila♦
308k33441775
asked Mar 14 at 13:34
Sean ThrasherSean Thrasher
444
edited Mar 14 at 19:47
Asaf Karagila♦
308k33441775
edited Mar 14 at 19:47
Asaf Karagila♦
308k33441775
edited Mar 14 at 19:47
Asaf Karagila♦
308k33441775
308k33441775
asked Mar 14 at 13:34
Sean ThrasherSean Thrasher
444
asked Mar 14 at 13:34
Sean ThrasherSean Thrasher
444
asked Mar 14 at 13:34
Sean ThrasherSean Thrasher
444
444
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^-1(x:x>0)$
answered Mar 14 at 13:40
Tsemo AristideTsemo Aristide
60.5k11446
$endgroup$
add a comment |
$begingroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus0$ with the usual distance and then $A=-1$, $B=1$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
answered Mar 14 at 13:40
Henning MakholmHenning Makholm
243k17311555
$endgroup$
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%2f3148005%2fprove-that-u-x-in-x-mid-dx-adx-b-is-open-when-a-and-b-are-disjoi%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^-1(x:x>0)$
answered Mar 14 at 13:40
Tsemo AristideTsemo Aristide
60.5k11446
$endgroup$
add a comment |
$begingroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^-1(x:x>0)$
answered Mar 14 at 13:40
Tsemo AristideTsemo Aristide
60.5k11446
$endgroup$
add a comment |
$begingroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^-1(x:x>0)$
answered Mar 14 at 13:40
Tsemo AristideTsemo Aristide
60.5k11446
$endgroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^-1(x:x>0)$
answered Mar 14 at 13:40
Tsemo AristideTsemo Aristide
60.5k11446
answered Mar 14 at 13:40
Tsemo AristideTsemo Aristide
60.5k11446
answered Mar 14 at 13:40
Tsemo AristideTsemo Aristide
60.5k11446
answered Mar 14 at 13:40
Tsemo AristideTsemo Aristide
60.5k11446
60.5k11446
add a comment |
add a comment |
$begingroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus0$ with the usual distance and then $A=-1$, $B=1$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
answered Mar 14 at 13:40
Henning MakholmHenning Makholm
243k17311555
$endgroup$
add a comment |
$begingroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus0$ with the usual distance and then $A=-1$, $B=1$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
answered Mar 14 at 13:40
Henning MakholmHenning Makholm
243k17311555
$endgroup$
add a comment |
$begingroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus0$ with the usual distance and then $A=-1$, $B=1$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
answered Mar 14 at 13:40
Henning MakholmHenning Makholm
243k17311555
$endgroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus0$ with the usual distance and then $A=-1$, $B=1$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
answered Mar 14 at 13:40
Henning MakholmHenning Makholm
243k17311555
edited Mar 14 at 13:49
answered Mar 14 at 13:40
Henning MakholmHenning Makholm
243k17311555
answered Mar 14 at 13:40
Henning MakholmHenning Makholm
243k17311555
answered Mar 14 at 13:40
Henning MakholmHenning Makholm
243k17311555
243k17311555
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%2f3148005%2fprove-that-u-x-in-x-mid-dx-adx-b-is-open-when-a-and-b-are-disjoi%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
