Why do closed-over variables survive after the outer function returns?
Because the inner function holds a reference to the outer lexical environment. As long as the inner function exists, the environment (and its variables) cannot be garbage collected. The scope chain persists.
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More FAQs in The Relationship Between Scope and Closures in JavaScript
Closures are a consequence of lexical scope. A function defined inside another keeps a reference to the outer lexical environment. When the outer function returns, the inner function still has access to those variables via the scope chain.
No. A closure is the combination of a function and its lexical environment (the variables it closes over). The function alone is not the closure; the function plus the retained outer variables is the closure.
No. Each call creates a separate lexical environment. So makeAdder(5) and makeAdder(10) each have their own x. The closures are independent.
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