Mixing
Neighbourhood, French railroad metric
$begingroup$
For every two elements $x,y$ of the disk $D^2$ let
$d(x,y)=begin{cases}|x-y|,text{if x and y are on the same line through (0,0)}\ |x|+|y|, text{else}end{cases}$
(the 'french-railroad-metric)
What do the neighborhoods of $(0,0)$ and $(frac12, 0)$ look like?
I want to figure out, what these neighborhoods look like.
For $x=(0,0)$ it is easy, because every $yin D^2$ can be on a line through (0,0).
So $d(x,y)=|x-y|=sqrt{y_1^2+y_2^2}<varepsilon$, which means that the neighborhoods of $(0,0)$ are just circles with centre (0,0) in $D^2$.
What do the neighborhoods of $x=(frac12,0)$ look like?
For every point on the x-axis we get $d(x,y)=sqrt{(y_1-frac12)^2+y_2^2}<varepsilon$.
Which are again circles with centre $(frac12, 0)$ on the disk.
If $y$ is not on the x-axis, we get:
$d(x,y)=|x|+|y|=frac12+|y|<varepsilon$
$|y|<varepsilon-frac12$.
Which are circles with centre (0,0) and a maximal radius of $1/2$?
So for $varepsilon=1$ for example, we would get this picture? (The red colored area)
Thanks in advance, for your thoughts.
metric-spaces
edited Jan 31 at 3:50
Parcly Taxel
44.7k1376109
asked Jan 31 at 3:35
CornmanCornman
3,36121229
$endgroup$
add a comment |
$begingroup$
For every two elements $x,y$ of the disk $D^2$ let
$d(x,y)=begin{cases}|x-y|,text{if x and y are on the same line through (0,0)}\ |x|+|y|, text{else}end{cases}$
(the 'french-railroad-metric)
What do the neighborhoods of $(0,0)$ and $(frac12, 0)$ look like?
I want to figure out, what these neighborhoods look like.
For $x=(0,0)$ it is easy, because every $yin D^2$ can be on a line through (0,0).
So $d(x,y)=|x-y|=sqrt{y_1^2+y_2^2}<varepsilon$, which means that the neighborhoods of $(0,0)$ are just circles with centre (0,0) in $D^2$.
What do the neighborhoods of $x=(frac12,0)$ look like?
For every point on the x-axis we get $d(x,y)=sqrt{(y_1-frac12)^2+y_2^2}<varepsilon$.
Which are again circles with centre $(frac12, 0)$ on the disk.
If $y$ is not on the x-axis, we get:
$d(x,y)=|x|+|y|=frac12+|y|<varepsilon$
$|y|<varepsilon-frac12$.
Which are circles with centre (0,0) and a maximal radius of $1/2$?
So for $varepsilon=1$ for example, we would get this picture? (The red colored area)
Thanks in advance, for your thoughts.
metric-spaces
edited Jan 31 at 3:50
Parcly Taxel
44.7k1376109
asked Jan 31 at 3:35
CornmanCornman
3,36121229
$endgroup$
add a comment |
$begingroup$
For every two elements $x,y$ of the disk $D^2$ let
$d(x,y)=begin{cases}|x-y|,text{if x and y are on the same line through (0,0)}\ |x|+|y|, text{else}end{cases}$
(the 'french-railroad-metric)
What do the neighborhoods of $(0,0)$ and $(frac12, 0)$ look like?
I want to figure out, what these neighborhoods look like.
For $x=(0,0)$ it is easy, because every $yin D^2$ can be on a line through (0,0).
So $d(x,y)=|x-y|=sqrt{y_1^2+y_2^2}<varepsilon$, which means that the neighborhoods of $(0,0)$ are just circles with centre (0,0) in $D^2$.
What do the neighborhoods of $x=(frac12,0)$ look like?
For every point on the x-axis we get $d(x,y)=sqrt{(y_1-frac12)^2+y_2^2}<varepsilon$.
Which are again circles with centre $(frac12, 0)$ on the disk.
If $y$ is not on the x-axis, we get:
$d(x,y)=|x|+|y|=frac12+|y|<varepsilon$
$|y|<varepsilon-frac12$.
Which are circles with centre (0,0) and a maximal radius of $1/2$?
So for $varepsilon=1$ for example, we would get this picture? (The red colored area)
Thanks in advance, for your thoughts.
metric-spaces
edited Jan 31 at 3:50
Parcly Taxel
44.7k1376109
asked Jan 31 at 3:35
CornmanCornman
3,36121229
$endgroup$
For every two elements $x,y$ of the disk $D^2$ let
$d(x,y)=begin{cases}|x-y|,text{if x and y are on the same line through (0,0)}\ |x|+|y|, text{else}end{cases}$
(the 'french-railroad-metric)
What do the neighborhoods of $(0,0)$ and $(frac12, 0)$ look like?
I want to figure out, what these neighborhoods look like.
For $x=(0,0)$ it is easy, because every $yin D^2$ can be on a line through (0,0).
So $d(x,y)=|x-y|=sqrt{y_1^2+y_2^2}<varepsilon$, which means that the neighborhoods of $(0,0)$ are just circles with centre (0,0) in $D^2$.
What do the neighborhoods of $x=(frac12,0)$ look like?
For every point on the x-axis we get $d(x,y)=sqrt{(y_1-frac12)^2+y_2^2}<varepsilon$.
Which are again circles with centre $(frac12, 0)$ on the disk.
If $y$ is not on the x-axis, we get:
$d(x,y)=|x|+|y|=frac12+|y|<varepsilon$
$|y|<varepsilon-frac12$.
Which are circles with centre (0,0) and a maximal radius of $1/2$?
So for $varepsilon=1$ for example, we would get this picture? (The red colored area)
Thanks in advance, for your thoughts.
metric-spaces
metric-spaces
edited Jan 31 at 3:50
Parcly Taxel
44.7k1376109
asked Jan 31 at 3:35
CornmanCornman
3,36121229
edited Jan 31 at 3:50
Parcly Taxel
44.7k1376109
asked Jan 31 at 3:35
CornmanCornman
3,36121229
edited Jan 31 at 3:50
Parcly Taxel
44.7k1376109
edited Jan 31 at 3:50
Parcly Taxel
44.7k1376109
edited Jan 31 at 3:50
Parcly Taxel
44.7k1376109
44.7k1376109
asked Jan 31 at 3:35
CornmanCornman
3,36121229
asked Jan 31 at 3:35
CornmanCornman
3,36121229
asked Jan 31 at 3:35
CornmanCornman
3,36121229
3,36121229
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, you wouldn't get that red-shaded area. The ball around $(frac12,0)$ of radius $1$ looks like
But that doesn't do a very good job of showing what's going on. We need a smaller ball for that. Here's the ball of radius $frac14$ centered at $(frac12,0)$, at the same scale:
If the radius is small enough that you can't reach Paris, you're stuck on the line you started on.
answered Jan 31 at 3:57
jmerryjmerry
17k11633
$endgroup$
$begingroup$
How do you get this red line? What have I done wrong? What I understand is, that if the condition $|y|<varepsilon-frac12$ 'fails', which means we get something negativ on the RHS (for $varepsilon=1/4$) we do not get a circle around (0,0). Why does the metric for points on the x-axis not give circles, but a straight line?
$endgroup$
– Cornman
Jan 31 at 4:05
$begingroup$
There are two cases in that distance function. The red line comes from the first case - staying on the same line. And if the radius is shorter than the distance to the origin, staying on that line is the only option.
$endgroup$
– jmerry
Jan 31 at 4:17
$begingroup$
Thank you. I see my mistake now. I had to set $y_2=0$ in my calculation, which then just gives a real number: $|y_1-frac12|$. Then it is clear, that we get this straight line and not cirlces!
$endgroup$
– Cornman
Jan 31 at 4:25
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%2f3094473%2fneighbourhood-french-railroad-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
$begingroup$
No, you wouldn't get that red-shaded area. The ball around $(frac12,0)$ of radius $1$ looks like
But that doesn't do a very good job of showing what's going on. We need a smaller ball for that. Here's the ball of radius $frac14$ centered at $(frac12,0)$, at the same scale:
If the radius is small enough that you can't reach Paris, you're stuck on the line you started on.
answered Jan 31 at 3:57
jmerryjmerry
17k11633
$endgroup$
$begingroup$
How do you get this red line? What have I done wrong? What I understand is, that if the condition $|y|<varepsilon-frac12$ 'fails', which means we get something negativ on the RHS (for $varepsilon=1/4$) we do not get a circle around (0,0). Why does the metric for points on the x-axis not give circles, but a straight line?
$endgroup$
– Cornman
Jan 31 at 4:05
$begingroup$
There are two cases in that distance function. The red line comes from the first case - staying on the same line. And if the radius is shorter than the distance to the origin, staying on that line is the only option.
$endgroup$
– jmerry
Jan 31 at 4:17
$begingroup$
Thank you. I see my mistake now. I had to set $y_2=0$ in my calculation, which then just gives a real number: $|y_1-frac12|$. Then it is clear, that we get this straight line and not cirlces!
$endgroup$
– Cornman
Jan 31 at 4:25
add a comment |
$begingroup$
No, you wouldn't get that red-shaded area. The ball around $(frac12,0)$ of radius $1$ looks like
But that doesn't do a very good job of showing what's going on. We need a smaller ball for that. Here's the ball of radius $frac14$ centered at $(frac12,0)$, at the same scale:
If the radius is small enough that you can't reach Paris, you're stuck on the line you started on.
answered Jan 31 at 3:57
jmerryjmerry
17k11633
$endgroup$
$begingroup$
How do you get this red line? What have I done wrong? What I understand is, that if the condition $|y|<varepsilon-frac12$ 'fails', which means we get something negativ on the RHS (for $varepsilon=1/4$) we do not get a circle around (0,0). Why does the metric for points on the x-axis not give circles, but a straight line?
$endgroup$
– Cornman
Jan 31 at 4:05
$begingroup$
There are two cases in that distance function. The red line comes from the first case - staying on the same line. And if the radius is shorter than the distance to the origin, staying on that line is the only option.
$endgroup$
– jmerry
Jan 31 at 4:17
$begingroup$
Thank you. I see my mistake now. I had to set $y_2=0$ in my calculation, which then just gives a real number: $|y_1-frac12|$. Then it is clear, that we get this straight line and not cirlces!
$endgroup$
– Cornman
Jan 31 at 4:25
add a comment |
$begingroup$
No, you wouldn't get that red-shaded area. The ball around $(frac12,0)$ of radius $1$ looks like
But that doesn't do a very good job of showing what's going on. We need a smaller ball for that. Here's the ball of radius $frac14$ centered at $(frac12,0)$, at the same scale:
If the radius is small enough that you can't reach Paris, you're stuck on the line you started on.
answered Jan 31 at 3:57
jmerryjmerry
17k11633
$endgroup$
No, you wouldn't get that red-shaded area. The ball around $(frac12,0)$ of radius $1$ looks like
But that doesn't do a very good job of showing what's going on. We need a smaller ball for that. Here's the ball of radius $frac14$ centered at $(frac12,0)$, at the same scale:
If the radius is small enough that you can't reach Paris, you're stuck on the line you started on.
answered Jan 31 at 3:57
jmerryjmerry
17k11633
answered Jan 31 at 3:57
jmerryjmerry
17k11633
answered Jan 31 at 3:57
jmerryjmerry
17k11633
answered Jan 31 at 3:57
jmerryjmerry
17k11633
17k11633
$begingroup$
How do you get this red line? What have I done wrong? What I understand is, that if the condition $|y|<varepsilon-frac12$ 'fails', which means we get something negativ on the RHS (for $varepsilon=1/4$) we do not get a circle around (0,0). Why does the metric for points on the x-axis not give circles, but a straight line?
$endgroup$
– Cornman
Jan 31 at 4:05
$begingroup$
There are two cases in that distance function. The red line comes from the first case - staying on the same line. And if the radius is shorter than the distance to the origin, staying on that line is the only option.
$endgroup$
– jmerry
Jan 31 at 4:17
$begingroup$
Thank you. I see my mistake now. I had to set $y_2=0$ in my calculation, which then just gives a real number: $|y_1-frac12|$. Then it is clear, that we get this straight line and not cirlces!
$endgroup$
– Cornman
Jan 31 at 4:25
add a comment |
$begingroup$
How do you get this red line? What have I done wrong? What I understand is, that if the condition $|y|<varepsilon-frac12$ 'fails', which means we get something negativ on the RHS (for $varepsilon=1/4$) we do not get a circle around (0,0). Why does the metric for points on the x-axis not give circles, but a straight line?
$endgroup$
– Cornman
Jan 31 at 4:05
$begingroup$
There are two cases in that distance function. The red line comes from the first case - staying on the same line. And if the radius is shorter than the distance to the origin, staying on that line is the only option.
$endgroup$
– jmerry
Jan 31 at 4:17
$begingroup$
Thank you. I see my mistake now. I had to set $y_2=0$ in my calculation, which then just gives a real number: $|y_1-frac12|$. Then it is clear, that we get this straight line and not cirlces!
$endgroup$
– Cornman
Jan 31 at 4:25
$begingroup$
How do you get this red line? What have I done wrong? What I understand is, that if the condition $|y|<varepsilon-frac12$ 'fails', which means we get something negativ on the RHS (for $varepsilon=1/4$) we do not get a circle around (0,0). Why does the metric for points on the x-axis not give circles, but a straight line?
$endgroup$
– Cornman
Jan 31 at 4:05
$begingroup$
How do you get this red line? What have I done wrong? What I understand is, that if the condition $|y|<varepsilon-frac12$ 'fails', which means we get something negativ on the RHS (for $varepsilon=1/4$) we do not get a circle around (0,0). Why does the metric for points on the x-axis not give circles, but a straight line?
$endgroup$
– Cornman
Jan 31 at 4:05
$begingroup$
There are two cases in that distance function. The red line comes from the first case - staying on the same line. And if the radius is shorter than the distance to the origin, staying on that line is the only option.
$endgroup$
– jmerry
Jan 31 at 4:17
$begingroup$
There are two cases in that distance function. The red line comes from the first case - staying on the same line. And if the radius is shorter than the distance to the origin, staying on that line is the only option.
$endgroup$
– jmerry
Jan 31 at 4:17
$begingroup$
Thank you. I see my mistake now. I had to set $y_2=0$ in my calculation, which then just gives a real number: $|y_1-frac12|$. Then it is clear, that we get this straight line and not cirlces!
$endgroup$
– Cornman
Jan 31 at 4:25
$begingroup$
Thank you. I see my mistake now. I had to set $y_2=0$ in my calculation, which then just gives a real number: $|y_1-frac12|$. Then it is clear, that we get this straight line and not cirlces!
$endgroup$
– Cornman
Jan 31 at 4:25
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%2f3094473%2fneighbourhood-french-railroad-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
