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A problem on path connectedness of unit ball in R*R

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$begingroup$

Actually, I am having a problem over the fact that as $f:[0,1]rightarrowmathbb{R}^{2}$ , $f(t)= (1-t)x + ty$ is continuous, $x,yinmathbb{R}^{2}$, then $f([0,1])$ is connected . Then as unit ball in $mathbb{R}^{2}$ is an ordered set in the order topology, by IVP of $f$ any point between $x=(0,1)$ and $y=(1,0)$ must have a pre-image in $[0,1]$ , but by the dictionary order topology, $(0,1)< z=(1/2,5/8) < (1,0)$ , but $z$ has no pre-image under $f$, but $z$ lies in the unit ball . Why ???
I want a clear explanation over this dubiousness .

general-topology connectedness path-connected

share|cite|improve this question

edited 17 hours ago

Robert Thingum

8931317

asked Mar 3 at 15:48

RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY

565

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is RR? Do you mean $mathbb{R}^2$? The unit ball is path connected, but you have to take paths $on the unit ball$. Think of the intersection of the ball with the plane that contains x, y, and the origin. The intersection is a great circle. The arc of the circle connecting x and y is a path.
  $endgroup$
  – Joel Pereira
  Mar 3 at 16:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes Joel, I mean R*R, my confusion lies in the fact that when the points of the unit ball are written on a line, then for f(0)=(0,1) and f(1)=(1,0) , then by dictionary order (1/2,5/8) lies in between f(0) and f(1), and it also lies on the unit ball , then by IVP of f, it must have a pre-image in [0,1], where is that ?????
  $endgroup$
  – RABI KUMAR CHAKRABORTY
  Mar 3 at 18:16
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Actually, I am having a problem over the fact that as $f:[0,1]rightarrowmathbb{R}^{2}$ , $f(t)= (1-t)x + ty$ is continuous, $x,yinmathbb{R}^{2}$, then $f([0,1])$ is connected . Then as unit ball in $mathbb{R}^{2}$ is an ordered set in the order topology, by IVP of $f$ any point between $x=(0,1)$ and $y=(1,0)$ must have a pre-image in $[0,1]$ , but by the dictionary order topology, $(0,1)< z=(1/2,5/8) < (1,0)$ , but $z$ has no pre-image under $f$, but $z$ lies in the unit ball . Why ???
I want a clear explanation over this dubiousness .

general-topology connectedness path-connected

share|cite|improve this question

edited 17 hours ago

Robert Thingum

8931317

asked Mar 3 at 15:48

RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY

565

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is RR? Do you mean $mathbb{R}^2$? The unit ball is path connected, but you have to take paths $on the unit ball$. Think of the intersection of the ball with the plane that contains x, y, and the origin. The intersection is a great circle. The arc of the circle connecting x and y is a path.
  $endgroup$
  – Joel Pereira
  Mar 3 at 16:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes Joel, I mean R*R, my confusion lies in the fact that when the points of the unit ball are written on a line, then for f(0)=(0,1) and f(1)=(1,0) , then by dictionary order (1/2,5/8) lies in between f(0) and f(1), and it also lies on the unit ball , then by IVP of f, it must have a pre-image in [0,1], where is that ?????
  $endgroup$
  – RABI KUMAR CHAKRABORTY
  Mar 3 at 18:16
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Actually, I am having a problem over the fact that as $f:[0,1]rightarrowmathbb{R}^{2}$ , $f(t)= (1-t)x + ty$ is continuous, $x,yinmathbb{R}^{2}$, then $f([0,1])$ is connected . Then as unit ball in $mathbb{R}^{2}$ is an ordered set in the order topology, by IVP of $f$ any point between $x=(0,1)$ and $y=(1,0)$ must have a pre-image in $[0,1]$ , but by the dictionary order topology, $(0,1)< z=(1/2,5/8) < (1,0)$ , but $z$ has no pre-image under $f$, but $z$ lies in the unit ball . Why ???
I want a clear explanation over this dubiousness .

general-topology connectedness path-connected

share|cite|improve this question

edited 17 hours ago

Robert Thingum

8931317

asked Mar 3 at 15:48

RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY

565

$endgroup$

share|cite|improve this question

edited 17 hours ago

Robert Thingum

8931317

asked Mar 3 at 15:48

RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY

565

share|cite|improve this question

edited 17 hours ago

Robert Thingum

8931317

asked Mar 3 at 15:48

RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY

565

edited 17 hours ago

Robert Thingum

8931317

edited 17 hours ago

Robert Thingum

8931317

asked Mar 3 at 15:48

RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY

565

asked Mar 3 at 15:48

RABI KUMAR CHAKRABORTYRABI KUMAR CHAKRABORTY

565