I am looking for a *constructive* proof of one of the following two statements. If they are not constructively provable, I would be very thankful for an explanation as to why that is so (i.e., at which point in a proof must non-constructive means be employed?).


1. There exists a normed space *X* such that for all *Y* $\subset$ *X*, if *Y* is denumerable, then *Y* is not dense in *X*.
2. There exists a normed space *X* such that for all *Y* $\subset$ *X*, if *Y* is dense in *X*, then *Y* is not denumerable.


I'd consider a proof constructive if it includes no applications of the:


* Law of Excluded Middle: $\phi$ $\lor$ $\neg$$\phi$
* Law of Double Negation Elimination: $\neg$$\neg$$\phi$ $\rightarrow$ $\phi$
* Axiom of Choice or any of its equivalents (Zorn's Lemma, etc.).


Suggestions would be much appreciated.



Constructive proof of existence of non-separable normed space
Andrew Swan and I proved that in the function realizability topos every metric space is separable (and that every object with decidable equality is countable). Therefore, it is not possible to prove constructively that a non-separable metric space exists. This result strengthens the answer by Matt Frank.


The note is available as [arxiv 1804.00427](https://arxiv.org/abs/1804.00427).


