As part of a larger question regarding proving a subset is not connected, so I have S = { ( x , y ) ∈ R 2 : x y = 1 } and I have said this is not connected as the function below: f ( x ) = { 1 x &gt; 0 0 x &lt; 0 is a surjective, continuous function that maps f : S → { 0 , 1 }So I now want to show f ( x ) is continuous, I have attempted to use l i m x → a f ( x ) = f ( a ) but I can't work out how to get this written out and what values of x I need to take? I'm also doubting whether this is continuous but for all values of x would not include 0 as it cannot take this value?

Question

As part of a larger question regarding proving a subset is not connected, so I have   S  =  {  (  x  ,  y  )  ∈            R        2    :  x  y  =  1  } and I have said this is not connected as the function below:  f  (  x  )  =      {                            1                          x          &amp;gt;          0                                      0                          x          &amp;lt;          0                        is a surjective, continuous function that maps   f  :  S  →  {  0  ,  1  }So I now want to show   f  (  x  ) is continuous, I have attempted to use   l  i      m          x      →      a        f  (  x  )  =  f  (  a  ) but I can&#039;t work out how to get this written out and what values of   x I need to take? I&#039;m also doubting whether this is continuous but for all values of   x would not include 0 as it cannot take this value?

Percyaehyq · Accepted Answer

Points in   S are pairs. So if you want to define a function   f  :  S  →     {  0  ,  1  }, you need to specify what the value of   f            (        (  x  ,  y  )            )       is. (We write shorter   f  (  x  ,  y  ).) Probably  f  (  (  x  ,  y  )  )  =      {                            1                          x          &amp;gt;          0                                      0                          x          &amp;lt;          0                        Now you want to show      lim          S      ∋      (      x      ,      y      )      →      (      a      ,      b      )        f  (  x  ,  y  )  =  f  (  a  ,  b  )for all   (  a  ,  b  )  ∈  S. So let   (      x    n    ,      y    n    ) a sequence converging to   (  a  ,  b  )  ∈  S. This implies that       x    n    →  a. If   a  =  0, then   a  b  =  0, so   (  a  ,  b  )  ∉  S. So   a  &amp;lt;  0 or   a  &amp;gt;  0. In both cases,       x    n   must have the same sign as   a for all sufficiantly big   n, so   f  (  (  x  ,  y  )  )  =  f  (  (  a  ,  b  )  ). Therefore       lim          (      x      ,      y      )      →             (      a      ,      b      )        f  (  x  ,  y  )  =  f  (  a  ,  b  ).   f is continous.

As part of a larger question regarding proving a subset is not connected, so I have S = <m

Answered question

Answer & Explanation

New Questions in Differential Calculus