Given f : [ 0 , 1 ] → R 2 such that f ( 0 ) = ( 1 , 0 ) , f ( 1 ) = ( − 1 , 0 ) and f is continuous with respect to the standard topologies. How can I prove that there exists x ∈ [ 0 , 1 ] such that f ( x ) = ( 0 , y ), for some y ∈ R ?My attempt to a solution was to consider the restriction of f to the set A = { x : f ( x ) = ( y , 0 ) for some y ∈ R } ⊂ [ 0 , 1 ] .Since f is continuous then its restriction is continuous.Since the restriction of f to A is "basically" a map from A to R , it makes sense that I would just need to apply IVT to finish the proof. I'm having issue in formalizing the reasoning that the restriction is "basically" a map from A to R .

Question

Given   f  :  [  0  ,  1  ]  →            R        2   such that   f  (  0  )  =  (  1  ,  0  )  ,  f  (  1  )  =  (  −  1  ,  0  ) and   f is continuous with respect to the standard topologies. How can I prove that there exists   x  ∈  [  0  ,  1  ] such that   f  (  x  )  =  (  0  ,  y  ), for some   y  ∈      R  ?My attempt to a solution was to consider the restriction of   f to the set  A  =  {  x  :  f  (  x  )  =  (  y  ,  0  )     for some     y  ∈      R    }  ⊂  [  0  ,  1  ]  .Since   f is continuous then its restriction is continuous.Since the restriction of   f to   A is &quot;basically&quot; a map from   A to       R  , it makes sense that I would just need to apply IVT to finish the proof. I&#039;m having issue in formalizing the reasoning that the restriction is &quot;basically&quot; a map from   A to       R  .

Andre Reynolds · Accepted Answer

Suppose such an   x does not exist. Then letting   D  =  {  (  0  ,  y  )  ,  y  ∈      R    }, we have   f  (  [  0  ,  1  ]  )  ⊂            R        2    ∖  D, which is a union of two disjoint open sets (namely the upper and lower half-plane), and hence disconnected. Then   f cannot be continuous or else   f  (  [  0  ,  1  ]  ) would be connected as the continuous image of a connected set.By the contrapositive, you have your claim.

Blaine Ortega · Accepted Answer

Attempt:Let       f    i  :   [  0  ,  1  ]  →      R  ,   i  =  1  ,  2.  f  =  (      f    1    ,      f    2    ) is continuos iff fi,   i  =  1  ,  2 are continuos.  f  (  x  )  =  (      f    1    (  x  )  ,      f    2    (  x  )  )Consider       f    1  :      f    1    (  0  )  =  1;       f    1    (  1  )  =  −  1,       f    1   continuos.IVT : There is a   p  ∈  [  0  ,  1  ] s.t.      f    1    (  p  )  =  0Let   y  :=      f    2    (  p  ), and we are done.

Given f:[0,1]->R^2 such that f(0)=(1,0),f(1)=(−1,0) and f is continuous with respect to the standard topologies. How can I prove that there exists x in [0,1] such that f(x)=(0,y), for some y in R?

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