Show that {(x,y) in R2 | xy<7} is open (closed ?) with respect to the l1 metric
$begingroup$
I really need your help.
I have to determine if {(x,y) in R2 | xy<7} is open or closed or both or neither, with respect to the Manhattan metric (or l1 metric or taxicab metric).
I have the feeling that it is open but I don’t manage to prove it.
I wanted to use the property : U open and f continuous <=> f-1(U) open, but I cannot prove that f is continuous with respect to the Manhattan metric.
Thank you very much
real-analysis analysis functions metric-spaces
edited Jan 24 at 13:33
daw
24.7k1645
asked Jan 24 at 11:08
user637846user637846
254
$endgroup$
add a comment |
$begingroup$
I really need your help.
I have to determine if {(x,y) in R2 | xy<7} is open or closed or both or neither, with respect to the Manhattan metric (or l1 metric or taxicab metric).
I have the feeling that it is open but I don’t manage to prove it.
I wanted to use the property : U open and f continuous <=> f-1(U) open, but I cannot prove that f is continuous with respect to the Manhattan metric.
Thank you very much
real-analysis analysis functions metric-spaces
edited Jan 24 at 13:33
daw
24.7k1645
asked Jan 24 at 11:08
user637846user637846
254
$endgroup$
add a comment |
$begingroup$
I really need your help.
I have to determine if {(x,y) in R2 | xy<7} is open or closed or both or neither, with respect to the Manhattan metric (or l1 metric or taxicab metric).
I have the feeling that it is open but I don’t manage to prove it.
I wanted to use the property : U open and f continuous <=> f-1(U) open, but I cannot prove that f is continuous with respect to the Manhattan metric.
Thank you very much
real-analysis analysis functions metric-spaces
edited Jan 24 at 13:33
daw
24.7k1645
asked Jan 24 at 11:08
user637846user637846
254
$endgroup$
I really need your help.
I have to determine if {(x,y) in R2 | xy<7} is open or closed or both or neither, with respect to the Manhattan metric (or l1 metric or taxicab metric).
I have the feeling that it is open but I don’t manage to prove it.
I wanted to use the property : U open and f continuous <=> f-1(U) open, but I cannot prove that f is continuous with respect to the Manhattan metric.
Thank you very much
real-analysis analysis functions metric-spaces
real-analysis analysis functions metric-spaces
edited Jan 24 at 13:33
daw
24.7k1645
asked Jan 24 at 11:08
user637846user637846
254
edited Jan 24 at 13:33
daw
24.7k1645
asked Jan 24 at 11:08
user637846user637846
254
edited Jan 24 at 13:33
daw
24.7k1645
edited Jan 24 at 13:33
daw
24.7k1645
edited Jan 24 at 13:33
daw
24.7k1645
24.7k1645
asked Jan 24 at 11:08
user637846user637846
254
asked Jan 24 at 11:08
user637846user637846
254
asked Jan 24 at 11:08
user637846user637846
254
254
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The topology on $mathbb R^2$ induced by the Manhattan metric coincides with the topology on $mathbb R^2$ induced by the usual Euclidean metric.
This follows directly from the inequalities $d_E(x,y)leq d_M(x,y)$ and $d_M(x,y)leq2d_E(x,y)$.
So in this context $mathbb R^2$ is equipped with its usual topology.
Further the function $f:mathbb R^2tomathbb R$ prescribed by $langle x,yranglemapsto xy$ is continuous and the set $(-infty,7)subseteqmathbb R$ is open.
We conclude that $f^{-1}((-infty,7))={langle x,yranglemid xy<7}$ is open.
The fact that $mathbb R^2$ is connected then excludes that the set is also closed.
answered Jan 24 at 11:25
drhabdrhab
103k545136
$endgroup$
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%2f3085748%2fshow-that-x-y-in-r2-xy7-is-open-closed-with-respect-to-the-l1-metric%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
$begingroup$
The topology on $mathbb R^2$ induced by the Manhattan metric coincides with the topology on $mathbb R^2$ induced by the usual Euclidean metric.
This follows directly from the inequalities $d_E(x,y)leq d_M(x,y)$ and $d_M(x,y)leq2d_E(x,y)$.
So in this context $mathbb R^2$ is equipped with its usual topology.
Further the function $f:mathbb R^2tomathbb R$ prescribed by $langle x,yranglemapsto xy$ is continuous and the set $(-infty,7)subseteqmathbb R$ is open.
We conclude that $f^{-1}((-infty,7))={langle x,yranglemid xy<7}$ is open.
The fact that $mathbb R^2$ is connected then excludes that the set is also closed.
answered Jan 24 at 11:25
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
The topology on $mathbb R^2$ induced by the Manhattan metric coincides with the topology on $mathbb R^2$ induced by the usual Euclidean metric.
This follows directly from the inequalities $d_E(x,y)leq d_M(x,y)$ and $d_M(x,y)leq2d_E(x,y)$.
So in this context $mathbb R^2$ is equipped with its usual topology.
Further the function $f:mathbb R^2tomathbb R$ prescribed by $langle x,yranglemapsto xy$ is continuous and the set $(-infty,7)subseteqmathbb R$ is open.
We conclude that $f^{-1}((-infty,7))={langle x,yranglemid xy<7}$ is open.
The fact that $mathbb R^2$ is connected then excludes that the set is also closed.
answered Jan 24 at 11:25
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
The topology on $mathbb R^2$ induced by the Manhattan metric coincides with the topology on $mathbb R^2$ induced by the usual Euclidean metric.
This follows directly from the inequalities $d_E(x,y)leq d_M(x,y)$ and $d_M(x,y)leq2d_E(x,y)$.
So in this context $mathbb R^2$ is equipped with its usual topology.
Further the function $f:mathbb R^2tomathbb R$ prescribed by $langle x,yranglemapsto xy$ is continuous and the set $(-infty,7)subseteqmathbb R$ is open.
We conclude that $f^{-1}((-infty,7))={langle x,yranglemid xy<7}$ is open.
The fact that $mathbb R^2$ is connected then excludes that the set is also closed.
answered Jan 24 at 11:25
drhabdrhab
103k545136
$endgroup$
The topology on $mathbb R^2$ induced by the Manhattan metric coincides with the topology on $mathbb R^2$ induced by the usual Euclidean metric.
This follows directly from the inequalities $d_E(x,y)leq d_M(x,y)$ and $d_M(x,y)leq2d_E(x,y)$.
So in this context $mathbb R^2$ is equipped with its usual topology.
Further the function $f:mathbb R^2tomathbb R$ prescribed by $langle x,yranglemapsto xy$ is continuous and the set $(-infty,7)subseteqmathbb R$ is open.
We conclude that $f^{-1}((-infty,7))={langle x,yranglemid xy<7}$ is open.
The fact that $mathbb R^2$ is connected then excludes that the set is also closed.
answered Jan 24 at 11:25
drhabdrhab
103k545136
edited Jan 24 at 11:53
answered Jan 24 at 11:25
drhabdrhab
103k545136
answered Jan 24 at 11:25
drhabdrhab
103k545136
answered Jan 24 at 11:25
drhabdrhab
103k545136
103k545136
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%2f3085748%2fshow-that-x-y-in-r2-xy7-is-open-closed-with-respect-to-the-l1-metric%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
