Check if function is continuous in metrix space
We have function $f:(mathbb{R^2}, d_{x}) rightarrow (mathbb{R^2}, d_{y})$ ($d_{x}$ - maximum metric space, $d_{y}$ - British rail metric space). $f(x,y) = (frac{x+y}{2}, frac{y-x}{2})$.
I need to check if $f$ and $f^{-1}$ are continuous.
$mathit{My, work}:$ for example, I took point $(2,2)$, and $f(2,2) = (2,0)$.
Here I used definition:
$forall varepsilon > 0 , exists delta >0: , max{{|x-2|,|y-2|}} < delta, , d_{y}(f(x,y), (2,0)) <varepsilon$. Let it be, $x=y$ and $varepsilon = 1$, so $|x-2| < delta$. Let $delta + 2 = x$. So, $d_{y}(f(delta+2, delta + 2), (2,0)) < varepsilon Rightarrow sqrt{(frac{4 + delta}{2} -2)^2 +(frac{2 + delta}{2})^2} = sqrt{frac{2delta^2 + 4delta + 4}{4}} > frac{delta + 2}{2} > 1$. So, $f$ couldnt be continuous. Is it right solution?
And here for $f^{-1}$: $left{begin{array}{l}a = frac{x+y}{2}\b= frac{y-x}{2}end{array}right. Rightarrow left{begin{array}{l} y= b+ a\x = a-bend{array}right. Rightarrow f^{-1} =(a-b, b+a)$ - is continuous?? I dont know how to show it.
general-topology
asked Nov 24 at 18:37
Aleksandra
565
add a comment |
We have function $f:(mathbb{R^2}, d_{x}) rightarrow (mathbb{R^2}, d_{y})$ ($d_{x}$ - maximum metric space, $d_{y}$ - British rail metric space). $f(x,y) = (frac{x+y}{2}, frac{y-x}{2})$.
I need to check if $f$ and $f^{-1}$ are continuous.
$mathit{My, work}:$ for example, I took point $(2,2)$, and $f(2,2) = (2,0)$.
Here I used definition:
$forall varepsilon > 0 , exists delta >0: , max{{|x-2|,|y-2|}} < delta, , d_{y}(f(x,y), (2,0)) <varepsilon$. Let it be, $x=y$ and $varepsilon = 1$, so $|x-2| < delta$. Let $delta + 2 = x$. So, $d_{y}(f(delta+2, delta + 2), (2,0)) < varepsilon Rightarrow sqrt{(frac{4 + delta}{2} -2)^2 +(frac{2 + delta}{2})^2} = sqrt{frac{2delta^2 + 4delta + 4}{4}} > frac{delta + 2}{2} > 1$. So, $f$ couldnt be continuous. Is it right solution?
And here for $f^{-1}$: $left{begin{array}{l}a = frac{x+y}{2}\b= frac{y-x}{2}end{array}right. Rightarrow left{begin{array}{l} y= b+ a\x = a-bend{array}right. Rightarrow f^{-1} =(a-b, b+a)$ - is continuous?? I dont know how to show it.
general-topology
asked Nov 24 at 18:37
Aleksandra
565
What is the British rail metric?
– Sean Roberson
Nov 24 at 18:49
@SeanRoberson it also called "post office metric" or "SNCF metric"
– Aleksandra
Nov 24 at 18:52
2
@Aleksandra Define it, not nickname it! We do not know what is in your head.
– Will M.
Nov 24 at 19:56
It looks like the SNCF metric is what is defined here: planetmath.org/sncfmetric
– Ingix
Nov 24 at 20:10
add a comment |
We have function $f:(mathbb{R^2}, d_{x}) rightarrow (mathbb{R^2}, d_{y})$ ($d_{x}$ - maximum metric space, $d_{y}$ - British rail metric space). $f(x,y) = (frac{x+y}{2}, frac{y-x}{2})$.
I need to check if $f$ and $f^{-1}$ are continuous.
$mathit{My, work}:$ for example, I took point $(2,2)$, and $f(2,2) = (2,0)$.
Here I used definition:
$forall varepsilon > 0 , exists delta >0: , max{{|x-2|,|y-2|}} < delta, , d_{y}(f(x,y), (2,0)) <varepsilon$. Let it be, $x=y$ and $varepsilon = 1$, so $|x-2| < delta$. Let $delta + 2 = x$. So, $d_{y}(f(delta+2, delta + 2), (2,0)) < varepsilon Rightarrow sqrt{(frac{4 + delta}{2} -2)^2 +(frac{2 + delta}{2})^2} = sqrt{frac{2delta^2 + 4delta + 4}{4}} > frac{delta + 2}{2} > 1$. So, $f$ couldnt be continuous. Is it right solution?
And here for $f^{-1}$: $left{begin{array}{l}a = frac{x+y}{2}\b= frac{y-x}{2}end{array}right. Rightarrow left{begin{array}{l} y= b+ a\x = a-bend{array}right. Rightarrow f^{-1} =(a-b, b+a)$ - is continuous?? I dont know how to show it.
general-topology
asked Nov 24 at 18:37
Aleksandra
565
We have function $f:(mathbb{R^2}, d_{x}) rightarrow (mathbb{R^2}, d_{y})$ ($d_{x}$ - maximum metric space, $d_{y}$ - British rail metric space). $f(x,y) = (frac{x+y}{2}, frac{y-x}{2})$.
I need to check if $f$ and $f^{-1}$ are continuous.
$mathit{My, work}:$ for example, I took point $(2,2)$, and $f(2,2) = (2,0)$.
Here I used definition:
$forall varepsilon > 0 , exists delta >0: , max{{|x-2|,|y-2|}} < delta, , d_{y}(f(x,y), (2,0)) <varepsilon$. Let it be, $x=y$ and $varepsilon = 1$, so $|x-2| < delta$. Let $delta + 2 = x$. So, $d_{y}(f(delta+2, delta + 2), (2,0)) < varepsilon Rightarrow sqrt{(frac{4 + delta}{2} -2)^2 +(frac{2 + delta}{2})^2} = sqrt{frac{2delta^2 + 4delta + 4}{4}} > frac{delta + 2}{2} > 1$. So, $f$ couldnt be continuous. Is it right solution?
And here for $f^{-1}$: $left{begin{array}{l}a = frac{x+y}{2}\b= frac{y-x}{2}end{array}right. Rightarrow left{begin{array}{l} y= b+ a\x = a-bend{array}right. Rightarrow f^{-1} =(a-b, b+a)$ - is continuous?? I dont know how to show it.
general-topology
general-topology
asked Nov 24 at 18:37
Aleksandra
565
asked Nov 24 at 18:37
Aleksandra
565
edited Nov 24 at 18:53
asked Nov 24 at 18:37
Aleksandra
565
asked Nov 24 at 18:37
Aleksandra
565
asked Nov 24 at 18:37
Aleksandra
565
565
What is the British rail metric?
– Sean Roberson
Nov 24 at 18:49
@SeanRoberson it also called "post office metric" or "SNCF metric"
– Aleksandra
Nov 24 at 18:52
2
@Aleksandra Define it, not nickname it! We do not know what is in your head.
– Will M.
Nov 24 at 19:56
It looks like the SNCF metric is what is defined here: planetmath.org/sncfmetric
– Ingix
Nov 24 at 20:10
add a comment |
What is the British rail metric?
– Sean Roberson
Nov 24 at 18:49
@SeanRoberson it also called "post office metric" or "SNCF metric"
– Aleksandra
Nov 24 at 18:52
2
@Aleksandra Define it, not nickname it! We do not know what is in your head.
– Will M.
Nov 24 at 19:56
It looks like the SNCF metric is what is defined here: planetmath.org/sncfmetric
– Ingix
Nov 24 at 20:10
What is the British rail metric?
– Sean Roberson
Nov 24 at 18:49
What is the British rail metric?
– Sean Roberson
Nov 24 at 18:49
@SeanRoberson it also called "post office metric" or "SNCF metric"
– Aleksandra
Nov 24 at 18:52
@SeanRoberson it also called "post office metric" or "SNCF metric"
– Aleksandra
Nov 24 at 18:52
2
2
@Aleksandra Define it, not nickname it! We do not know what is in your head.
– Will M.
Nov 24 at 19:56
@Aleksandra Define it, not nickname it! We do not know what is in your head.
– Will M.
Nov 24 at 19:56
It looks like the SNCF metric is what is defined here: planetmath.org/sncfmetric
– Ingix
Nov 24 at 20:10
It looks like the SNCF metric is what is defined here: planetmath.org/sncfmetric
– Ingix
Nov 24 at 20:10
add a comment |
1 Answer
1
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oldest
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In $mathbb{R}^2$ in the $d_y$ metric, all singleton sets except ${0,0)}$ are open, and $(0,0)$ has essentially the same neighbourhoods as in the usual metric on the plane. The max metric $d_x$ is just equivalent to the usual Euclidean metric.
$f$ is a bijection, sending $(0,0)$ to $(0,0)$ and as $f$ is usual-usual continuous, it is usual-$d_y$ continuous at $(0,0)$ as well. It is not continuous at any other point, because the usual topology (or the max metric) has no isolated points at all. The inverse function $f^{-1}$ is continuous at all points unequal to $(0,0)$ because those points are isolated in $d_y$, and similar reasoning as above will give that $f^{-1}$ is continuous at $(0,0)$ as well.
So $f^{-1}$ is everywhere continuous and $f$ only at the origin.
answered Nov 24 at 20:37
Henno Brandsma
105k346113
add a comment |
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In $mathbb{R}^2$ in the $d_y$ metric, all singleton sets except ${0,0)}$ are open, and $(0,0)$ has essentially the same neighbourhoods as in the usual metric on the plane. The max metric $d_x$ is just equivalent to the usual Euclidean metric.
$f$ is a bijection, sending $(0,0)$ to $(0,0)$ and as $f$ is usual-usual continuous, it is usual-$d_y$ continuous at $(0,0)$ as well. It is not continuous at any other point, because the usual topology (or the max metric) has no isolated points at all. The inverse function $f^{-1}$ is continuous at all points unequal to $(0,0)$ because those points are isolated in $d_y$, and similar reasoning as above will give that $f^{-1}$ is continuous at $(0,0)$ as well.
So $f^{-1}$ is everywhere continuous and $f$ only at the origin.
answered Nov 24 at 20:37
Henno Brandsma
105k346113
add a comment |
In $mathbb{R}^2$ in the $d_y$ metric, all singleton sets except ${0,0)}$ are open, and $(0,0)$ has essentially the same neighbourhoods as in the usual metric on the plane. The max metric $d_x$ is just equivalent to the usual Euclidean metric.
$f$ is a bijection, sending $(0,0)$ to $(0,0)$ and as $f$ is usual-usual continuous, it is usual-$d_y$ continuous at $(0,0)$ as well. It is not continuous at any other point, because the usual topology (or the max metric) has no isolated points at all. The inverse function $f^{-1}$ is continuous at all points unequal to $(0,0)$ because those points are isolated in $d_y$, and similar reasoning as above will give that $f^{-1}$ is continuous at $(0,0)$ as well.
So $f^{-1}$ is everywhere continuous and $f$ only at the origin.
answered Nov 24 at 20:37
Henno Brandsma
105k346113
add a comment |
In $mathbb{R}^2$ in the $d_y$ metric, all singleton sets except ${0,0)}$ are open, and $(0,0)$ has essentially the same neighbourhoods as in the usual metric on the plane. The max metric $d_x$ is just equivalent to the usual Euclidean metric.
$f$ is a bijection, sending $(0,0)$ to $(0,0)$ and as $f$ is usual-usual continuous, it is usual-$d_y$ continuous at $(0,0)$ as well. It is not continuous at any other point, because the usual topology (or the max metric) has no isolated points at all. The inverse function $f^{-1}$ is continuous at all points unequal to $(0,0)$ because those points are isolated in $d_y$, and similar reasoning as above will give that $f^{-1}$ is continuous at $(0,0)$ as well.
So $f^{-1}$ is everywhere continuous and $f$ only at the origin.
answered Nov 24 at 20:37
Henno Brandsma
105k346113
In $mathbb{R}^2$ in the $d_y$ metric, all singleton sets except ${0,0)}$ are open, and $(0,0)$ has essentially the same neighbourhoods as in the usual metric on the plane. The max metric $d_x$ is just equivalent to the usual Euclidean metric.
$f$ is a bijection, sending $(0,0)$ to $(0,0)$ and as $f$ is usual-usual continuous, it is usual-$d_y$ continuous at $(0,0)$ as well. It is not continuous at any other point, because the usual topology (or the max metric) has no isolated points at all. The inverse function $f^{-1}$ is continuous at all points unequal to $(0,0)$ because those points are isolated in $d_y$, and similar reasoning as above will give that $f^{-1}$ is continuous at $(0,0)$ as well.
So $f^{-1}$ is everywhere continuous and $f$ only at the origin.
answered Nov 24 at 20:37
Henno Brandsma
105k346113
answered Nov 24 at 20:37
Henno Brandsma
105k346113
answered Nov 24 at 20:37
Henno Brandsma
105k346113
answered Nov 24 at 20:37
Henno Brandsma
105k346113
105k346113
add a comment |
add a comment |
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What is the British rail metric?
– Sean Roberson
Nov 24 at 18:49
@SeanRoberson it also called "post office metric" or "SNCF metric"
– Aleksandra
Nov 24 at 18:52
2
@Aleksandra Define it, not nickname it! We do not know what is in your head.
– Will M.
Nov 24 at 19:56
It looks like the SNCF metric is what is defined here: planetmath.org/sncfmetric
– Ingix
Nov 24 at 20:10