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@ Henriksen

The functions attain the value 1 on a rational for all but finitely many 'n' . Hence these would not give us continuous functions taking (only) rationals to infinity.

@ Gerry

I'm guessing I did not clearly understand the construction of the functions.

If 'x' is an irrational of the first kind (you stated) then why would the functions assume the value 0 on 'x' for all but finitely many 'n' ? I believe the 'n'th function is constructed by defining it to be the piecewise join of points,some with the 2nd co-ordinate being 'n' and some with the 2nd co-ordinate being '0' but the first co-ordinate being a rational number for all the points. The reals where the nth function was defined to be 0 were exactly at a distance of n^(-4) from a rational with denominator less than or equal to 'n'. If this is the case then the functions would never assume the value '0' on any irrational.

2 added 726 characters in body

@ Henriksen

The functions attain the value 1 on a rational for all but finitely many 'n' . Hence these would not give us continuous functions taking (only) rationals to infinity.

@ Gerry

I'm guessing I did not clearly understand the construction of the functions.

If 'x' is an irrational of the first kind (you stated) then why would the functions assume the value 0 on 'x' for all but finitely many 'n' ? I believe the 'n'th function is constructed by defining it to be the piecewise join of points,some with the 2nd co-ordinate being 'n' and some with the 2nd co-ordinate being '0' but the first co-ordinate being a rational number for all the points. The reals where the nth function was defined to be 0 were exactly at a distance of n^(-4) from a rational with denominator less than or equal to 'n'. If this is the case then the functions would never assume the value '0' on any irrational.

1

@ Henriksen

The functions attain the value 1 on a rational for all but finitely many 'n' . Hence these would not give us continuous functions taking (only) rationals to infinity.