Re: [math-fun] Can a nonconstant continuous function on the reals

8 Feb 2016

      I'll bite on the first question: No.

Suppose 

     f: R —> R 

is a nonconstant continuous function taking only rational values, say at least the values p < q in Q.

Let c be an irrational number such that

        p  <  c  <  q
.

Let finv(X) denote the inverse image by f of the set X.

Then by assumption we have

   R  =  finv((-oo,c))  +  finv((c,oo))

where + denotes disjoint union.

This exhibits the connected set R of reals as the disjoint union of two nonempty open sets — contradiction.

—Dan
...
On Feb 7, 2016, at 10:08 PM, Bill Gosper <billgosper@gmail.com> wrote:
assume only rational values?  Related puzzle:
Find an x in closed form s.t. CantorStaircase[x] is irrational.
--rwg
Spoiler:
"2/)]2/1 ,0 ,3[atehTcitpillE + 1-( == ]]3/1 ,0 ,3[atehTcitpillE +
1-[esacriatSrotnaC"

Re: [math-fun] Can a nonconstant continuous function on the reals

Dan Asimov