Distance function from x to the Cantor set

  • Context: Graduate 
  • Thread starter Thread starter Dragonfall
  • Start date Start date
  • Tags Tags
    Cantor Function Set
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 3K views
Dragonfall
Messages
1,023
Reaction score
5
Does the said function satisfy:

(1)continuity
(2)never constant
(3)has uncountably many zeroes

1 and 3 is trivial, but I'm not sure about 2.
 
Physics news on Phys.org
CRGreathouse said:
I'm not sure how you'd even define the distance, given that every point in (0, 1) has a point in the Cantor set within epsilon for any epsilon > 0.

Not true. The Cantor set is not dense in (0, 1). For example the point 1/2 is at least 1/6 away from any point in the Cantor (middle-thirds) set. In fact, dist({1/2}, CantorSet} = 1/6.
 
Dragonfall said:
Does the said function satisfy:

(1)continuity
(2)never constant
(3)has uncountably many zeroes

1 and 3 is trivial, but I'm not sure about 2.

If x=1/2, then the distance from x to the Cantor middle-third set would be 1/6. If x=0, then the distance would be 0. Hence "not constant".

I find the the use of the word "never" strange since it sounds to be like asserting otherwise the function would be constant on, say, the Tuesdays after a new moon, but not constant all other days.

Possibly what you mean is that there are no open sets on which the function is constant.
 
By "never" I mean that there is no interval on which it is constant.

This is was a problem I thought up. My intuition was that since if a function is "continuous", and "never constant", each time you hit a zero you must "wave" up and down in order to hit a zero again. So this will make the number of zeros "countable". But the distance function from x to the cantor set seems to be "continuous and never constant" but has uncountably many zeros.