Theorem: Let ( X , d X ) and Y ( d Y ) be metric spaces. A function f : X → Y is continuous if and only if for every open U ⊂ Y, then f − 1 ( U ) = { x ∈ X : f ( x ) ∈ U }is open in X.If I want to show that the function f : [ 0 , 1 ] ∪ [ 2 , 4 ] ⊂ R → R defined by f ( x ) = { 1 , x ∈ [ 0 , 1 ] ; 2 , x ∈ [ 2 , 4 ] . is continuous, then I need to consider any open set U ⊂ R and show that f − 1 ( U ) = { x ∈ [ 0 , 1 ] ∪ [ 2 , 4 ] : f ( x ) ∈ U }is open.I understand that if 1 ∉ U and 2 ∉ U, then f − 1 ( U ) = ∅. But I don't understand what happens when only 1 ∈ U, or only 2 ∈ U, or 1 , 2 ∈ U.If only 1 ∈ U, then is it correct to say that f − 1 ( U )=[0,1]? If only 2 ∈ U, then is it correct to say that f − 1 ( U )=[2,4]? But I am really confused since these are closed intervals. I am lost.Can someone help me to find f − 1 ( U ) for those cases? Any clues or hints will be appreciated.

Question

Theorem: Let   (  X  ,      d    X    ) and   Y  (      d    Y    ) be metric spaces. A function   f  :  X  →  Y is continuous if and only if for every open   U  ⊂  Y, then      f          −      1        (  U  )  =  {  x  ∈  X  :  f  (  x  )  ∈  U  }is open in   X.If I want to show that the function   f  :  [  0  ,  1  ]  ∪  [  2  ,  4  ]  ⊂      R    →      R   defined by  f  (  x  )  =      {                            1          ,                          x          ∈          [          0          ,          1          ]          ;                                      2          ,                          x          ∈          [          2          ,          4          ]          .                        is continuous, then I need to consider any open set   U  ⊂      R   and show that      f          −      1        (  U  )  =  {  x  ∈  [  0  ,  1  ]  ∪  [  2  ,  4  ]  :  f  (  x  )  ∈  U  }is open.I understand that if   1  ∉  U and   2  ∉  U, then       f          −      1        (  U  )  =  ∅. But I don&#039;t understand what happens when only   1  ∈  U, or only   2  ∈  U, or   1  ,  2  ∈  U.If only   1  ∈  U, then is it correct to say that       f          −      1        (  U  )=[0,1]? If only   2  ∈  U, then is it correct to say that       f          −      1        (  U  )=[2,4]? But I am really confused since these are closed intervals. I am lost.Can someone help me to find       f          −      1        (  U  ) for those cases? Any clues or hints will be appreciated.

Kyler Crawford · Accepted Answer

You are missing the fact that   [  1  ,  2  ] and   [  2  ,  4  ] are actually open sets in   [  0  ,  1  ]  ∪  [  2  ,  4  ]. For example,   [  0  ,  1  ]  =  {  x  ∈  [  0  ,  1  ]  ∪  [  2  ,  4  ]  :      |    x  −  0      |    &amp;lt;  1.5  }.So   1  ∈  U  ,  2  ∉  U gives       f          −      1        (  U  )  =  [  0  ,  1  ] which is open.

Theorem: Let ( X , d X </msub> ) and Y ( d Y

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