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Is ${0}$ is open in $(mathbb{R}^2 , p)$?

How to describe the family $tau$ of all open sets of $(mathbb R^2,delta)$open & closed sets in $mathbb{R}$ and $mathbb{R}^2$Is A = ${(0,0)}cup {(x,sin(1/x))|0 < x le 1}$ $subseteq mathbb{R}^2$ under the usual metric on $mathbb{R}$ is compact?show that f is homomorphism?the set $U ={(x,-y) in mathbb{R}^2 : x=0,1,-1text{ and }y in mathbb{N}}$ is which of the following statement is True?In usual Euclidean metric on $mathbb{R}^n$. Which of the following metric spaces X is completeId $: mathbb{R} rightarrow mathbb{ R}$ is the identity mapping then choose the correct statementProperties of dictionary order topologyFinding the sphere?Is $d$ a metric on $X$?

0

$begingroup$

Associate with $mathbb{R}^2 $ the euclidean metric and denote its by $d$ . Define on $mathbb{R}^2 times mathbb{R}^2 $ a function $p$ by
$$p(x,y) = begin{cases} 0, text{x=y} \ d(x,0) + d(y,0) , x neq y end{cases}$$ where $0 = (0,0)$

Now my question is that

Is ${0}$ is open in $(mathbb{R}^2 , p)$?

My attempt : I thinks yes because we have $B(a,r) = { x in mathbb{R}^2 : d(x,0) + d(a,0) < r}$ and $B(0,r) subset {0}$

Is its true ?

Any hints/solution will be appreciated

thanks u

general-topology metric-spaces

share|cite|improve this question

edited Mar 19 at 3:55

jasmine

asked Mar 19 at 3:52

jasminejasmine

1,941420

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So you are using the $mathcal{L}_1$ metric. Under any $mathcal{L}_n$ metric, finite sets are not open.
  $endgroup$
  – Don Thousand
  Mar 19 at 3:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $B(a,r) = { x in mathbb{R}^2 : p(a,x) < r}.$ So, in particular, $B(0,r) = { x in mathbb{R}^2 : p(0,x) < r} = { x in mathbb{R}^2 : d(0,x) < r}.$ Clearly, there are other $x in mathbb{R}^2$ than $0$ that belongs to $B(0,r),$ such as $x=(0,r/2)$ for example
  $endgroup$
  – Pebeto
  Mar 19 at 4:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonThousand Is this not more like the notorious French railway metric? Except that it is not quite that either. Here the train from $x$ to Paris has no intermediate stops.
  $endgroup$
  – Jyrki Lahtonen
  Mar 19 at 4:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen Perhaps I misread? I shall look again.
  $endgroup$
  – Don Thousand
  Mar 19 at 4:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think B (0,r) $subset $ {0}? That's obviously false p((r/3,r/3),(0,0))=2r/3 <r. So (r/3,r/3) in B (0,r).
  $endgroup$
  – fleablood
  Mar 19 at 5:51
  

  
  
  
  

add a comment | 

0

$begingroup$

Associate with $mathbb{R}^2 $ the euclidean metric and denote its by $d$ . Define on $mathbb{R}^2 times mathbb{R}^2 $ a function $p$ by
$$p(x,y) = begin{cases} 0, text{x=y} \ d(x,0) + d(y,0) , x neq y end{cases}$$ where $0 = (0,0)$

Now my question is that

Is ${0}$ is open in $(mathbb{R}^2 , p)$?

My attempt : I thinks yes because we have $B(a,r) = { x in mathbb{R}^2 : d(x,0) + d(a,0) < r}$ and $B(0,r) subset {0}$

Is its true ?

Any hints/solution will be appreciated

thanks u

general-topology metric-spaces

share|cite|improve this question

edited Mar 19 at 3:55

jasmine

asked Mar 19 at 3:52

jasminejasmine

1,941420

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So you are using the $mathcal{L}_1$ metric. Under any $mathcal{L}_n$ metric, finite sets are not open.
  $endgroup$
  – Don Thousand
  Mar 19 at 3:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $B(a,r) = { x in mathbb{R}^2 : p(a,x) < r}.$ So, in particular, $B(0,r) = { x in mathbb{R}^2 : p(0,x) < r} = { x in mathbb{R}^2 : d(0,x) < r}.$ Clearly, there are other $x in mathbb{R}^2$ than $0$ that belongs to $B(0,r),$ such as $x=(0,r/2)$ for example
  $endgroup$
  – Pebeto
  Mar 19 at 4:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonThousand Is this not more like the notorious French railway metric? Except that it is not quite that either. Here the train from $x$ to Paris has no intermediate stops.
  $endgroup$
  – Jyrki Lahtonen
  Mar 19 at 4:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen Perhaps I misread? I shall look again.
  $endgroup$
  – Don Thousand
  Mar 19 at 4:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think B (0,r) $subset $ {0}? That's obviously false p((r/3,r/3),(0,0))=2r/3 <r. So (r/3,r/3) in B (0,r).
  $endgroup$
  – fleablood
  Mar 19 at 5:51
  

  
  
  
  

add a comment | 

$begingroup$

Associate with $mathbb{R}^2 $ the euclidean metric and denote its by $d$ . Define on $mathbb{R}^2 times mathbb{R}^2 $ a function $p$ by
$$p(x,y) = begin{cases} 0, text{x=y} \ d(x,0) + d(y,0) , x neq y end{cases}$$ where $0 = (0,0)$

Now my question is that

Is ${0}$ is open in $(mathbb{R}^2 , p)$?

My attempt : I thinks yes because we have $B(a,r) = { x in mathbb{R}^2 : d(x,0) + d(a,0) < r}$ and $B(0,r) subset {0}$

Is its true ?

Any hints/solution will be appreciated

thanks u

general-topology metric-spaces

share|cite|improve this question

edited Mar 19 at 3:55

jasmine

asked Mar 19 at 3:52

jasminejasmine

1,941420

$endgroup$

share|cite|improve this question

edited Mar 19 at 3:55

jasmine

asked Mar 19 at 3:52

jasminejasmine

1,941420

share|cite|improve this question

edited Mar 19 at 3:55

jasmine

asked Mar 19 at 3:52

jasminejasmine

1,941420

asked Mar 19 at 3:52

jasminejasmine

1,941420

asked Mar 19 at 3:52

jasminejasmine

1,941420

2 Answers
2

active

oldest

votes

1

$begingroup$

No. because $B(0,r) = { x : d(0,x) < r }$

is not a subset of ${0}$ because $(0,frac{r}{2})$ in $B(0,r).$

share|cite|improve this answer

edited Mar 19 at 4:32

jasmine

1,941420

answered Mar 19 at 4:15

William ElliotWilliam Elliot

8,9562820

$endgroup$

add a comment | 

2

$begingroup$

If ${0}$ is open, then there exists $r>0$ such that ${xin mathbb{R}^2:p(x,0)<r}subseteq {0}$, i.e., ${xin mathbb{R}^2:d(x,0)<r}subseteq{0}$, which is not possible as $d$ is the Euclidean norm.( For example, $(frac{r}{2sqrt{2}},frac{r}{2sqrt{2}})$ is in ${xin mathbb{R}^2:d(x,0)<r}$)

share|cite|improve this answer

answered Mar 19 at 4:24

SurajitSurajit

64819

$endgroup$

add a comment | 

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1

$begingroup$

No. because $B(0,r) = { x : d(0,x) < r }$

is not a subset of ${0}$ because $(0,frac{r}{2})$ in $B(0,r).$

share|cite|improve this answer

edited Mar 19 at 4:32

jasmine

1,941420

answered Mar 19 at 4:15

William ElliotWilliam Elliot

8,9562820

$endgroup$

add a comment | 

1

$begingroup$

No. because $B(0,r) = { x : d(0,x) < r }$

is not a subset of ${0}$ because $(0,frac{r}{2})$ in $B(0,r).$

share|cite|improve this answer

edited Mar 19 at 4:32

jasmine

1,941420

answered Mar 19 at 4:15

William ElliotWilliam Elliot

8,9562820

$endgroup$

add a comment | 

$begingroup$

No. because $B(0,r) = { x : d(0,x) < r }$

is not a subset of ${0}$ because $(0,frac{r}{2})$ in $B(0,r).$

share|cite|improve this answer

edited Mar 19 at 4:32

jasmine

1,941420

answered Mar 19 at 4:15

William ElliotWilliam Elliot

8,9562820

$endgroup$

share|cite|improve this answer

edited Mar 19 at 4:32

jasmine

1,941420

answered Mar 19 at 4:15

William ElliotWilliam Elliot

8,9562820

edited Mar 19 at 4:32

jasmine

1,941420

edited Mar 19 at 4:32

jasmine

1,941420

answered Mar 19 at 4:15

William ElliotWilliam Elliot

8,9562820

answered Mar 19 at 4:15

William ElliotWilliam Elliot

8,9562820

2

$begingroup$

If ${0}$ is open, then there exists $r>0$ such that ${xin mathbb{R}^2:p(x,0)<r}subseteq {0}$, i.e., ${xin mathbb{R}^2:d(x,0)<r}subseteq{0}$, which is not possible as $d$ is the Euclidean norm.( For example, $(frac{r}{2sqrt{2}},frac{r}{2sqrt{2}})$ is in ${xin mathbb{R}^2:d(x,0)<r}$)

share|cite|improve this answer

answered Mar 19 at 4:24

SurajitSurajit

64819

$endgroup$

add a comment | 

2

$begingroup$

If ${0}$ is open, then there exists $r>0$ such that ${xin mathbb{R}^2:p(x,0)<r}subseteq {0}$, i.e., ${xin mathbb{R}^2:d(x,0)<r}subseteq{0}$, which is not possible as $d$ is the Euclidean norm.( For example, $(frac{r}{2sqrt{2}},frac{r}{2sqrt{2}})$ is in ${xin mathbb{R}^2:d(x,0)<r}$)

share|cite|improve this answer

answered Mar 19 at 4:24

SurajitSurajit

64819

$endgroup$

add a comment | 

$begingroup$

If ${0}$ is open, then there exists $r>0$ such that ${xin mathbb{R}^2:p(x,0)<r}subseteq {0}$, i.e., ${xin mathbb{R}^2:d(x,0)<r}subseteq{0}$, which is not possible as $d$ is the Euclidean norm.( For example, $(frac{r}{2sqrt{2}},frac{r}{2sqrt{2}})$ is in ${xin mathbb{R}^2:d(x,0)<r}$)

share|cite|improve this answer

answered Mar 19 at 4:24

SurajitSurajit

64819

$endgroup$

share|cite|improve this answer

answered Mar 19 at 4:24

SurajitSurajit

64819

answered Mar 19 at 4:24

SurajitSurajit

64819

answered Mar 19 at 4:24

SurajitSurajit

64819