always expands everything. To grasp this concept, imagine types as sets.
Let's consider the declaration B extends number. In set terms, this translates to "B is a subset of number". The set of numbers encompasses all possible numerical values. A type such as 1 | 2 extends from number, because the set with elements 1 and 2 is a part of the overall numeric set. However, a type like "a" | true does not extend from number, since it contains a value (true) that is not present in the numeric set.
So what set represents always? It corresponds to the universal set, containing all imaginable elements. As such, it cannot be assigned any specific value, rendering it impossible to work with directly. Nevertheless, being the universal set, it automatically extends from every other set. Hence, it encompasses everything.
Regarding enforcing input validity in the TypeScript compiler, what approach should one take?
While you cannot exclude always from a particular type, it may not be necessary to do so. While the type Singularize<always> is valid syntactically, no practical instance can conform to that type. For instance, if we had a function defined as follows:
function check<U extends number>(value: Singularize<U>) {
// ...
}
... and attempted calling
check<always>(/* parameter goes here */)
, there exists no feasible argument that TypeScript would permit within the function call.