Discussion on: Sets in JavaScript — the only explanation you will ever need

[Alex Lohr]

Your examples can be simplified by using arrays as intermediate step:

const isSuperset = (set, superset) => [...set].every((item) => superset.has(item))
const union = (set1, set2) => new Set([...set1, ...set2])
const intersection = (set1, set2) => new Set([...set1].filter((item) => set2.has(item)))
const symmetricDifference = (set1, set2) => new Set([...set1, ...set2].filter((item) => set1.has(item) ^ set2.has(item)))
const difference = (set1, set2) => new Set([...set1].filter((item) => !set2.has(item)))

[Marcos Molina 👩🏽‍💻]

Thank you, just added to the post.