Define a list of optional keys for Typescript Record

You can create the partial version of your List type:

type PartialList = Partial<List>;

And you could do it all on one line if you don't want the intermediate type:

type PartialList = Partial<Record<'a' | 'b' | 'c', string>>;

You might decide that, in the end, the most expressive for your future self is:

type List = {
    a?: string;
    b?: string;
    c?: string;
}

Looks like in new versions of typescript you may do the following

type YourUnion = 'a' | 'b' | 'c';   
type ObjectWithOptionalKeys = Partial<Record<YourUnion, string>>
const someObject: ObjectWithOptionalKeys {
  a: 'str', // works
  b: 1 // throws
}
// c may not be specified at all

There is no way to specify the optionality of members of Record. They are required by definition

type Record<K extends keyof any, T> = {
    [P in K]: T; // Mapped properties are not optional, and it's not a homomorphic mapped type so it can't come from anywhere else.
};

You can define your own type if this is a common scenario for you:

type PartialRecord<K extends keyof any, T> = {
  [P in K]?: T;
};
type List =  PartialRecord<'a' | 'b' | 'c', string>

Or you can define PartialRecord using the predefined mapped types as well:

type PartialRecord<K extends keyof any, T> =  Partial<Record<K, T>>