Measure, Integration & Real Analysis · Problem 6D.21
L^p Spaces — Sheldon Axler
Mathematics — Measure Theory · convex-subset-dense-complement-dense
Problem
Suppose is an infinite-dimensional normed vector space. Show that there is a convex subset of such that and such that the complement is also a convex subset of with . [See 8.25 for the definition of a convex set. This exercise should stretch your geometric intuition because this behavior cannot happen in finite dimensions.]Intuitive explanation
In finite dimensions, convex sets with convex complements are rare and rigid—think of a hyperplane dividing space. But infinite-dimensional spaces have "room to wiggle": you can construct two interlocking convex sets that are both dense and whose complements are also convex, something impossible when the space has finite dimension. The key is to exploit the lack of a finite basis and build using a carefully chosen linear functional or sequence that creates a "thin" separating structure that's dense in both directions.
Intuitive explanation
In finite dimensions, convex sets with convex complements are rare and rigid—think of a hyperplane dividing space. But infinite-dimensional spaces have "room to wiggle": you can construct two interlocking convex sets that are both dense and whose complements are also convex, something impossible when the space has finite dimension. The key is to exploit the lack of a finite basis and build using a carefully chosen linear functional or sequence that creates a "thin" separating structure that's dense in both directions.