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Showing content from http://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set/delete below:

Set.prototype.delete() - JavaScript | MDN

Set.prototype.delete()

Baseline Widely available

The delete() method of Set instances removes a specified value from this set, if it is in the set.

Try it
const set1 = new Set();
set1.add({ x: 10, y: 20 }).add({ x: 20, y: 30 });

// Delete any point with `x > 10`.
set1.forEach((point) => {
  if (point.x > 10) {
    set1.delete(point);
  }
});

console.log(set1.size);
// Expected output: 1
Syntax
setInstance.delete(value)
Parameters
value

The value to remove from Set.

Return value

Returns true if value was already in Set; otherwise false.

Examples Using the delete() method
const mySet = new Set();
mySet.add("foo");

console.log(mySet.delete("bar")); // false; no "bar" element found to be deleted.
console.log(mySet.delete("foo")); // true; successfully removed.

console.log(mySet.has("foo")); // false; the "foo" element is no longer present.
Deleting an object from a set

Because objects are compared by reference, you have to delete them by checking individual properties if you don't have a reference to the original object.

const setObj = new Set(); // Create a new set.

setObj.add({ x: 10, y: 20 }); // Add object in the set.

setObj.add({ x: 20, y: 30 }); // Add object in the set.

// Delete any point with `x > 10`.
setObj.forEach((point) => {
  if (point.x > 10) {
    setObj.delete(point);
  }
});
Specifications Browser compatibility See also

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