Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ such that for all non-empty, open, and uncountable subset $E \subseteq \mathbb{R}$, $f(E) = \mathbb{R}$?

I have been thinking about it for a bit, but really can't think of any way to prove or disprove it. I have a hunch that if a set $E \subseteq \mathbb{R}$ is open and uncountable, then there should be a bijection to $\mathbb{R}$. This may not be true, however. I'm not well versed in the more advanced concepts of cardinality, so I'm unsure if there is an uncountable set $E \subsetneq \mathbb{R}$ such that there is no bijection from $E$ to $\mathbb R$.

I'm having great difficulty imagining a function with this property. I understand that proving existence doesn't imply that construction of such a function is feasible, but I want to understand if such a function even exists.

Any help would be greatly appreciated.


marked as duplicate by Ross Millikan, C. Falcon, Daniel W. Farlow, Noble Mushtak, Shailesh Jan 1 '17 at 0:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Darn! I searched for quite a bit and couldn't find it. Nice find. Yes, it's pretty much a duplicate. I guess his function implies mine exists, so it would be sufficient. $\endgroup$ – Enrico Borba Dec 31 '16 at 17:11
  • $\begingroup$ I also asked a very similar question here. $\endgroup$ – Arthur Dec 31 '16 at 17:13
  • $\begingroup$ I gave an explicit construction in my answer to the linked question. You still need that every open set includes an interval. $\endgroup$ – Ross Millikan Dec 31 '16 at 17:15

Yes. The Conway Base 13 is a function $f:\mathbb{R} \to \mathbb{R}$ such that its restriction to any open interval is surjective.


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