(b) Show that if |J|>|P(ℕ)|, then ℝJ does not contain a countable dense subset.
Proof: (a) For every oddly finite sequence q1,...,q2n+1 of rationals such that 0≤q2<q4<...<q2n≤1, let there be an associated function I→ℝ that takes value q2k−1 on (q2k−2,q2k) for each k=1,...,n (where q0=0), takes value q2n+1 on (q2n,1], and is zero on each of q2,...,q2n. We see the collection of such functions is countable, and we claim it forms a countable dense subset of ℝI: For a given basis element ∏i∈IUi of open sets Ui⊆ℝ such that Ui≠ℝ for only finitely many i∈I, let i1<...<in be exactly all such that Uij≠ℝ. We may choose rationals q2,...,q2(n−1) such that i1<q2<i2<q4<...<q2(n−1)<in (unless n=1 where one chooses i1<q2 and say q3=0) and for each q1,q3,...,q2n−1 choose rationals such that q2k−1∈Uik for each k=1,...,n. Then the associated function on q1,...,q2n−1 takes a value within Ui on each i, and hence is contained in the basis element.
(b) Suppose D is a countable subset of ℝJ. Fix some nonempty interval (a,b), and choose some disjoint nonempty interval (c,d). Then since the function f:J→D given by f(α)=D∩π−1α(a,b) cannot be injective, we find some distinct α,β∈ℝ such that for each d∈D we have d(α)∈(a,b) iff d(β)∈(a,b). But now the basis element ∏Uj where Uα=(a,b) and Uβ=(c,d) and Uj=ℝ otherwise cannot contain a point of D. ◻
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