It is not completely clear how [Bridges, Richman and Youchuan](http://www.sciencedirect.com/science/article/pii/S0019357797891318) treated sets in their paper. Example is in the following lemma (Lemma 7 on p. 7):



> 
> Let $U$ and $V$ be (**inhabited** to mean $\exists u \in U, \exists v \in V$) sets of a Banach space such that $U \cup V$ is dense. Then,
> the following holds:
> 
> 
> 1. If $u\_0 \in U$ and $v\_0 \in V$, then $\rho([u\_0, v\_0], \bar U \cap \bar V) = 0$
> 2. $\rho(x, \bar U \cap \bar V) = \rho(x, U) \wedge \rho(x, V)$
> 
> 
> ...
> 
> 
> To do this, choose $w$ in $U \cup V$ within ...
> 
> 
> 


If a set $X$ is inhabited, then there is (at least) **one** point that can be constructed. How can **other** points be constructed?



Sets in constructive mathematics
**The lemma has an existence hypothesis; it is this existence hypothesis that allows the proof to “choose” a point when necessary.**


$\newcommand{\x}{\vec{x}}$Specifically, the lemma assumes that $U \cup V$ is dense. That says (by definition) that for every suitable $x$ and $\varepsilon$, there exists some point in $U \cup V$ within $\epsilon$ of $x$.


But then the logical rules for the existential quantifier tell us: if we’re trying to prove some goal, and we have established that there exists some object with some property, it suffices to prove the goal assuming that we have some *specific* object with that property.


Formally, this is the natural deduction rule that says: If $\Gamma,\, \varphi(\x, y) \vdash \psi(\x)$, then $\Gamma,\, \exists y. \varphi(\x, y)\vdash \psi(\x)$ (where $y$ is not free in $\Gamma$ or $\psi(\x)$).
(Most other formalisations of constructive logic have some similar rule or axiom.) But in prose, it is often phrased as e.g. “We know there exists some $y$ such that […]. So, choose some such $y$; then …”


So the proof is not “constructing” some new point in the sense which — as you say — would be impossible in general. It is just choosing (for the purpose of the proof in question) points whose existence is guaranteed by the assumption that $U \cup V$ is dense.


