Mixing
Exist a norm such that $S= {x= (x_{1},x_{2}) in R^{2}| |x| leq 1, x_{2} >0}$ which isn't open nor closed?
$begingroup$
Let $X = R^{2}$ and $|.|$ a norm in X:
Show that exist a norm such that $S= {x= (x_{1},x_{2}) in R^{2}| |x| leq 1, x_{2} >0}$ it is not open or closed.
How can I show this?
Any help?
real-analysis general-topology norm normed-spaces
edited Jan 21 at 7:37
Martin Sleziak
44.8k10119272
asked Jan 20 at 18:23
LinkmanLinkman
1416
$endgroup$
|
show 1 more comment
$begingroup$
Let $X = R^{2}$ and $|.|$ a norm in X:
Show that exist a norm such that $S= {x= (x_{1},x_{2}) in R^{2}| |x| leq 1, x_{2} >0}$ it is not open or closed.
How can I show this?
Any help?
real-analysis general-topology norm normed-spaces
edited Jan 21 at 7:37
Martin Sleziak
44.8k10119272
asked Jan 20 at 18:23
LinkmanLinkman
1416
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– Calvin Khor
Jan 20 at 18:26
$begingroup$
@CalvinKhor thanks for your attention. I can show that for $||x|| leq 1$ in $R$ it is not open and closed. But it left me confused when the set is on $R^{2}$.
$endgroup$
– Linkman
Jan 20 at 18:30
$begingroup$
The set of real numbers $x$ such that $|x|le 1 $ is not open.
$endgroup$
– Calvin Khor
Jan 20 at 18:31
$begingroup$
@CalvinKhor I edited.
$endgroup$
– Linkman
Jan 20 at 18:32
$begingroup$
I see, sorry. Perhaps as a warm-up you can show that ${ x in mathbb R : |x|le 1 , x>0}$ is neither open or closed. And then just try to do the original question with your favourite norm on $mathbb R^2$.
$endgroup$
– Calvin Khor
Jan 20 at 18:34
|
show 1 more comment
$begingroup$
Let $X = R^{2}$ and $|.|$ a norm in X:
Show that exist a norm such that $S= {x= (x_{1},x_{2}) in R^{2}| |x| leq 1, x_{2} >0}$ it is not open or closed.
How can I show this?
Any help?
real-analysis general-topology norm normed-spaces
edited Jan 21 at 7:37
Martin Sleziak
44.8k10119272
asked Jan 20 at 18:23
LinkmanLinkman
1416
$endgroup$
Let $X = R^{2}$ and $|.|$ a norm in X:
Show that exist a norm such that $S= {x= (x_{1},x_{2}) in R^{2}| |x| leq 1, x_{2} >0}$ it is not open or closed.
How can I show this?
Any help?
real-analysis general-topology norm normed-spaces
real-analysis general-topology norm normed-spaces
edited Jan 21 at 7:37
Martin Sleziak
44.8k10119272
asked Jan 20 at 18:23
LinkmanLinkman
1416
edited Jan 21 at 7:37
Martin Sleziak
44.8k10119272
asked Jan 20 at 18:23
LinkmanLinkman
1416
edited Jan 21 at 7:37
Martin Sleziak
44.8k10119272
edited Jan 21 at 7:37
Martin Sleziak
44.8k10119272
edited Jan 21 at 7:37
Martin Sleziak
44.8k10119272
44.8k10119272
asked Jan 20 at 18:23
LinkmanLinkman
1416
asked Jan 20 at 18:23
LinkmanLinkman
1416
asked Jan 20 at 18:23
LinkmanLinkman
1416
1416
$begingroup$
What have you tried?
$endgroup$
– Calvin Khor
Jan 20 at 18:26
$begingroup$
@CalvinKhor thanks for your attention. I can show that for $||x|| leq 1$ in $R$ it is not open and closed. But it left me confused when the set is on $R^{2}$.
$endgroup$
– Linkman
Jan 20 at 18:30
$begingroup$
The set of real numbers $x$ such that $|x|le 1 $ is not open.
$endgroup$
– Calvin Khor
Jan 20 at 18:31
$begingroup$
@CalvinKhor I edited.
$endgroup$
– Linkman
Jan 20 at 18:32
$begingroup$
I see, sorry. Perhaps as a warm-up you can show that ${ x in mathbb R : |x|le 1 , x>0}$ is neither open or closed. And then just try to do the original question with your favourite norm on $mathbb R^2$.
$endgroup$
– Calvin Khor
Jan 20 at 18:34
|
show 1 more comment
$begingroup$
What have you tried?
$endgroup$
– Calvin Khor
Jan 20 at 18:26
$begingroup$
@CalvinKhor thanks for your attention. I can show that for $||x|| leq 1$ in $R$ it is not open and closed. But it left me confused when the set is on $R^{2}$.
$endgroup$
– Linkman
Jan 20 at 18:30
$begingroup$
The set of real numbers $x$ such that $|x|le 1 $ is not open.
$endgroup$
– Calvin Khor
Jan 20 at 18:31
$begingroup$
@CalvinKhor I edited.
$endgroup$
– Linkman
Jan 20 at 18:32
$begingroup$
I see, sorry. Perhaps as a warm-up you can show that ${ x in mathbb R : |x|le 1 , x>0}$ is neither open or closed. And then just try to do the original question with your favourite norm on $mathbb R^2$.
$endgroup$
– Calvin Khor
Jan 20 at 18:34
$begingroup$
What have you tried?
$endgroup$
– Calvin Khor
Jan 20 at 18:26
$begingroup$
What have you tried?
$endgroup$
– Calvin Khor
Jan 20 at 18:26
$begingroup$
@CalvinKhor thanks for your attention. I can show that for $||x|| leq 1$ in $R$ it is not open and closed. But it left me confused when the set is on $R^{2}$.
$endgroup$
– Linkman
Jan 20 at 18:30
$begingroup$
@CalvinKhor thanks for your attention. I can show that for $||x|| leq 1$ in $R$ it is not open and closed. But it left me confused when the set is on $R^{2}$.
$endgroup$
– Linkman
Jan 20 at 18:30
$begingroup$
The set of real numbers $x$ such that $|x|le 1 $ is not open.
$endgroup$
– Calvin Khor
Jan 20 at 18:31
$begingroup$
The set of real numbers $x$ such that $|x|le 1 $ is not open.
$endgroup$
– Calvin Khor
Jan 20 at 18:31
$begingroup$
@CalvinKhor I edited.
$endgroup$
– Linkman
Jan 20 at 18:32
$begingroup$
@CalvinKhor I edited.
$endgroup$
– Linkman
Jan 20 at 18:32
$begingroup$
I see, sorry. Perhaps as a warm-up you can show that ${ x in mathbb R : |x|le 1 , x>0}$ is neither open or closed. And then just try to do the original question with your favourite norm on $mathbb R^2$.
$endgroup$
– Calvin Khor
Jan 20 at 18:34
$begingroup$
I see, sorry. Perhaps as a warm-up you can show that ${ x in mathbb R : |x|le 1 , x>0}$ is neither open or closed. And then just try to do the original question with your favourite norm on $mathbb R^2$.
$endgroup$
– Calvin Khor
Jan 20 at 18:34
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
This is somewhat of a trick question: For every norm $|cdot|$ on $Bbb R^2$, the set
$$S={,x=(x_1,x_2)inBbb R^2mid|x|le 1,x_2>0,} $$
is neither open nor closed (under the topology induced by the norm).
In order to see that $S$ is not open, note that $(0,1)ne(0,0)$ implies $|(0,1)|>0$, hence for $a:=(0,frac1{|(0,1)|})$, we have $|a|=1$. For all $epsilon>0$, the open ball ${,xinBbb R^2mid|x-a|<epsilon,}$ fails to be a subset of $S$ because for $x:=(1+fracepsilon2)a$ we check that $|x-a|=|fracepsilon2a|=fracepsilon2<epsilon$ and $|x|=1+fracepsilon2>1$.
In order to see that $S$ is not closed, i.e., that the complement of $S$ is not open, note that $(0,0)notin S$, but every open ball ${,xinBbb R^2mid |x|<epsilon,}$, $epsilon>0$, intersects $S$. Indeed, we only need to look at $fracepsilon2a$ with $a$ as defined above.
answered Jan 21 at 7:21
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
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%2f3080949%2fexist-a-norm-such-that-s-x-x-1-x-2-in-r2-x-leq-1-x-2-0%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
$begingroup$
This is somewhat of a trick question: For every norm $|cdot|$ on $Bbb R^2$, the set
$$S={,x=(x_1,x_2)inBbb R^2mid|x|le 1,x_2>0,} $$
is neither open nor closed (under the topology induced by the norm).
In order to see that $S$ is not open, note that $(0,1)ne(0,0)$ implies $|(0,1)|>0$, hence for $a:=(0,frac1{|(0,1)|})$, we have $|a|=1$. For all $epsilon>0$, the open ball ${,xinBbb R^2mid|x-a|<epsilon,}$ fails to be a subset of $S$ because for $x:=(1+fracepsilon2)a$ we check that $|x-a|=|fracepsilon2a|=fracepsilon2<epsilon$ and $|x|=1+fracepsilon2>1$.
In order to see that $S$ is not closed, i.e., that the complement of $S$ is not open, note that $(0,0)notin S$, but every open ball ${,xinBbb R^2mid |x|<epsilon,}$, $epsilon>0$, intersects $S$. Indeed, we only need to look at $fracepsilon2a$ with $a$ as defined above.
answered Jan 21 at 7:21
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
add a comment |
$begingroup$
This is somewhat of a trick question: For every norm $|cdot|$ on $Bbb R^2$, the set
$$S={,x=(x_1,x_2)inBbb R^2mid|x|le 1,x_2>0,} $$
is neither open nor closed (under the topology induced by the norm).
In order to see that $S$ is not open, note that $(0,1)ne(0,0)$ implies $|(0,1)|>0$, hence for $a:=(0,frac1{|(0,1)|})$, we have $|a|=1$. For all $epsilon>0$, the open ball ${,xinBbb R^2mid|x-a|<epsilon,}$ fails to be a subset of $S$ because for $x:=(1+fracepsilon2)a$ we check that $|x-a|=|fracepsilon2a|=fracepsilon2<epsilon$ and $|x|=1+fracepsilon2>1$.
In order to see that $S$ is not closed, i.e., that the complement of $S$ is not open, note that $(0,0)notin S$, but every open ball ${,xinBbb R^2mid |x|<epsilon,}$, $epsilon>0$, intersects $S$. Indeed, we only need to look at $fracepsilon2a$ with $a$ as defined above.
answered Jan 21 at 7:21
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
add a comment |
$begingroup$
This is somewhat of a trick question: For every norm $|cdot|$ on $Bbb R^2$, the set
$$S={,x=(x_1,x_2)inBbb R^2mid|x|le 1,x_2>0,} $$
is neither open nor closed (under the topology induced by the norm).
In order to see that $S$ is not open, note that $(0,1)ne(0,0)$ implies $|(0,1)|>0$, hence for $a:=(0,frac1{|(0,1)|})$, we have $|a|=1$. For all $epsilon>0$, the open ball ${,xinBbb R^2mid|x-a|<epsilon,}$ fails to be a subset of $S$ because for $x:=(1+fracepsilon2)a$ we check that $|x-a|=|fracepsilon2a|=fracepsilon2<epsilon$ and $|x|=1+fracepsilon2>1$.
In order to see that $S$ is not closed, i.e., that the complement of $S$ is not open, note that $(0,0)notin S$, but every open ball ${,xinBbb R^2mid |x|<epsilon,}$, $epsilon>0$, intersects $S$. Indeed, we only need to look at $fracepsilon2a$ with $a$ as defined above.
answered Jan 21 at 7:21
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
This is somewhat of a trick question: For every norm $|cdot|$ on $Bbb R^2$, the set
$$S={,x=(x_1,x_2)inBbb R^2mid|x|le 1,x_2>0,} $$
is neither open nor closed (under the topology induced by the norm).
In order to see that $S$ is not open, note that $(0,1)ne(0,0)$ implies $|(0,1)|>0$, hence for $a:=(0,frac1{|(0,1)|})$, we have $|a|=1$. For all $epsilon>0$, the open ball ${,xinBbb R^2mid|x-a|<epsilon,}$ fails to be a subset of $S$ because for $x:=(1+fracepsilon2)a$ we check that $|x-a|=|fracepsilon2a|=fracepsilon2<epsilon$ and $|x|=1+fracepsilon2>1$.
In order to see that $S$ is not closed, i.e., that the complement of $S$ is not open, note that $(0,0)notin S$, but every open ball ${,xinBbb R^2mid |x|<epsilon,}$, $epsilon>0$, intersects $S$. Indeed, we only need to look at $fracepsilon2a$ with $a$ as defined above.
answered Jan 21 at 7:21
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 21 at 7:21
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 21 at 7:21
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 21 at 7:21
Hagen von EitzenHagen von Eitzen
282k23272505
282k23272505
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%2f3080949%2fexist-a-norm-such-that-s-x-x-1-x-2-in-r2-x-leq-1-x-2-0%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
$begingroup$
What have you tried?
$endgroup$
– Calvin Khor
Jan 20 at 18:26
$begingroup$
@CalvinKhor thanks for your attention. I can show that for $||x|| leq 1$ in $R$ it is not open and closed. But it left me confused when the set is on $R^{2}$.
$endgroup$
– Linkman
Jan 20 at 18:30
$begingroup$
The set of real numbers $x$ such that $|x|le 1 $ is not open.
$endgroup$
– Calvin Khor
Jan 20 at 18:31
$begingroup$
@CalvinKhor I edited.
$endgroup$
– Linkman
Jan 20 at 18:32
$begingroup$
I see, sorry. Perhaps as a warm-up you can show that ${ x in mathbb R : |x|le 1 , x>0}$ is neither open or closed. And then just try to do the original question with your favourite norm on $mathbb R^2$.
$endgroup$
– Calvin Khor
Jan 20 at 18:34