Let f : R → R be a continuous map such as each irrational number is mapped to a rational number (i.e. f ( R ∖ Q ) ⊂ Q ). Show that f is a constant map.

Question

Let   f  :      R    →      R   be a continuous map such as each irrational number is mapped to a rational number (i.e.  f  (      R    ∖      Q    )  ⊂      Q  ). Show that   f is a constant map.

Deanna Sweeney · Accepted Answer

Let&#039;s suppose that   f is not a constant map, i.e. it exists   x  ,  y  ∈      R   such that   f  (  x  )  ≠  f  (  y  )  .. As   f is continuous, the intermediate value theorem gives us that   f  (      R    ∖      Q    )  ⊂      Q  . But, as  f  (  [  x  ,  y  ]  )  ⊂  f  (      R    )  =  f  (      R    ∖      Q      ∪        Q    )  =  f  (      R    ∖      Q    )    ∪        ⋃          n      ∈              N              {  f  (      p    n    )  }  ⊂      Q      ∪        ⋃          n      ∈              N              {  f  (      p    n    )  }  ,where       p    n   is a sequence which describes       Q    ,, we would get that   [  f  (  x  )  ,  f  (  y  )  ] is a subset of a countable set and so is countable, which is a contradiction and so   f is constant.

Let f:R->R be a continuous map such as each irrational number is mapped to a rational number (i.e. f(R∖Q)sub Q). Show that f is a constant map.

Answered question

Answer & Explanation

New Questions in Algebra I