For each x ∈ [ 0 , 10 ), write its decimal expansion x = a 0 . a 1 a 2 a 3 … a n …. Take into account the number x ′ formed by the digits of odd rank of this decimal expansion; i. e., x ′ = 0. a 1 a 3 a 5 … a 2 n − 1 … We have two cases:(1) x ′ is not periodic(2) x ′ is periodic after the digit a ( 2 n 0 − 1 ) Now let f be the function from the half-open interval [0,10) to R defined by f ( x ) = { 0 in case (1) for x ′ a ( 2 n 0 ) . a 2 ( n 0 + 1 ) … a 2 ( n 0 + k ) … in case (2) for x ′ (Note that in case (2), the integer part [ f ( x ) ] = a ( 2 n 0 ) of f ( x ) can be nonzero.)QUESTIONProve that for all x 1 , x 2 , the function f takes on the open interval ( x 1 , x 2 ) all values between f ( x 1 ) and f ( x 2 ) despite f being discontinuous in all points of its domain.

Question

For each   x  ∈  [  0  ,  10  ), write its decimal expansion   x  =      a    0    .      a    1        a    2        a    3    …      a    n    …. Take into account the number       x    ′   formed by the digits of odd rank of this decimal expansion; i. e.,       x    ′    =  0.      a    1        a    3        a    5    …      a          2      n      −      1        … We have two cases:(1)       x    ′   is not periodic(2)       x    ′   is periodic after the digit       a          (      2              n        0            −      1      )      Now let   f be the function from the half-open interval [0,10) to       R   defined by  f  (  x  )  =      {                            0                          in case (1) for                       x            ′                                                            a                          (              2                              n                0                            )                                .                      a                          2              (                              n                0                            +              1              )                                …                      a                          2              (                              n                0                            +              k              )                                …                          in case (2) for                       x            ′                                  (Note that in case (2), the integer part   [  f  (  x  )  ]  =      a          (      2              n        0            )       of   f  (  x  ) can be nonzero.)QUESTIONProve that for all       x    1    ,      x    2  , the function   f takes on the open interval   (      x    1    ,      x    2    ) all values between   f  (      x    1    ) and   f  (      x    2    ) despite   f being discontinuous in all points of its domain.

dtal50 · Accepted Answer

There is an even stronger property than requiblack: On any inhabited open interval   (      x    1    ,      x    2    ) the function f takes all values in the interval   [  0  ,  10  ). As both   f  (      x    1    ) and   f  (      x    2    )  ∈  [  0  ,  10  ), this property is stronger than your question.First we pick a suitable subset of   (      x    1    ,      x    2    ) to make think nicer. There is some number   t  =      t    0    .      t    1        t    2        t    3    …      t          2      n      +      1       such that both   t and   t  +      10          −      (      2      n      +      1      )       are in   (      x    1    ,      x    2    ) and       t          2      n      +      1       is neither 0 nor 9.To get any desiblack value   y  =      b    0    .      b    1        b    2        b    3    … just construct       x    y    =      t    0    .      t    1        t    2    …      t          2      n      +      1            b    0    0      b    1    0      b    2    0  …. Note that we are in case   (  2  ) with       n    0    =  n.There are probbably some issues for numbers without a unique decimal expansion – but your definition already ignores those.

Answered question

Answer & Explanation

New Questions in Calculus 2