How come I didn't have to instantiate the NSCharacterSet object first before using it?

I'm following a tutorial learning about delegates and protocols while using the UITextField object. In the tutorial I noticed that I didn't have to instantiate the NSCharacterSet object first by typing

let letterCharacters = NSCharacterSet()

The tutorial's code worked by reaching directly for the letters variable of NSCharacterSet which is mind blowing. I assumed objects always needed to be instantiated first before using them or making reference to them. Here is the full function that works:

func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
        let existingTextHasDecimalSeperator = textField.text?.range(of: ".")
        let replacementTextHasDecimalSeperator = string.range(of: ".")
        let letterCharacters = NSCharacterSet.letters
        let foundCharacter = string.rangeOfCharacter(from: letterCharacters)


        if existingTextHasDecimalSeperator != nil && replacementTextHasDecimalSeperator != nil {
            return false
        } else if foundCharacter != nil {
            return false
        } else {
            return true
        }
    }

Why am I allowed to use NSCharacterSet.letters directly instead of creating a NSCharacterSet object first?

Solution

From the Apple documentation:

class var letters: CharacterSet

A character set containing the characters in Unicode General Category L* & M*.

The key piece here is the class part of that definition. If it is a class variable, than you do not need to instantiate the object with NSCharacterSet().

A quick primer on variable types.

A class variable can be accessed simply by calling the variable with the class name CharacterSet.letters for example
An instance variable needs to have its class instantiated first before it can be accessed, characterSet.inverted would be an example of this

Typically, a class variable would never change so it has no need to maintain state. An instance variable likely will.