From the jQuery docs javascript guide:

Because local scope works through functions, any functions defined within another have access to variables defined in the outer function:

function outer() {
    var x = 5;
    var y = 2;
    function inner() {
        console.log( x );
        debugger; // <-- !!!
    }
    inner();
}
outer()

Console triggered with debugger:

> x
5
> y
ReferenceError: y is not defined

Since variables defined in the outer function can be used by the inner function (E.g. x or y), why is the debugger not able to call the y variable?

I suspect people will answer that the debugger only shows variables defined in the most inner/local scope. The reason for this being that otherwise no distinction could be made using the debugger between the inner and outer scope when inspecting a variable using the debugger in the inner function. Additionally, every variable defined in an outer scope which is executed in the inner scope allows the debugger to access it.

But if that is the case, isn't there some way to still call the variable y from the console inside the inner function? (using a notation respectful of scope, e.g. outer.y)

Edit: Debuggers in other languages

Apparently this behavior of a debugger is not limited to javascript. The Python debugger pdb for example behaves similarly:

def outer():
    x = 5
    y = 2
    def inner():
        print x
        import pdb; pdb.set_trace()
    inner()
outer()

(Pdb) x
5
(Pdb) y
*** NameError: 'y' is not defined