> they are much more careful than the average bear to explicitly qualify inaccurate things as inaccurate
Sure. But what part of this entirely worded in natural language, and very short statement made you think it was a technical, formal statement? I think you’re just taking an opportunity to flex your knowledge of basic calculus, and deliberately attributing intent to the author that isn’t there in order to look clever.
Regarding a function being discontinuous at a point outside its domain: if you take a completely naive view of what ‘discontinuous’ means, then I suppose you can say so. But discontinuity is just the logical negation of continuity. Observe:
To say that f: X —> Y (in this context, a real-valued function of real numbers) is continuous means precisely
∀x∈X ∀ε>0 ∃δ>0 |x - p| < δ ⇒ |f(x) - f(p)| < ε
and so its negation looks like
∃x∈X ⌐ …
that is, there is a point in X, the domain of f where continuity fails.
For example, you wouldn’t talk about a function defined on the integers being discontinuous at pi, would you? That would just be weird.
To prove the point further, observe that the set of discontinuities (according to your definition) of any given function would actually include every number… in fact every mathematical object in the universe — which would make it not even a set in ZFC. So it’s absurd.
Even more reasons to believe functions can only be discontinuous at points of their domain: a function is said to be discontinuous if it has at least one discontinuity. By your definition, every function is discontinuous.
…anyway, I said we were going to be petty. I’m trying to demonstrate this is a waste of time by wasting my own time.
you have an interesting point of view, and some of the things you have said are correct, but if you try to use gradient descent on a function from, say, ℤ → ℝ, you are going to be a very sad xanda. i would indeed describe such a function as being discontinuous not just at π but everywhere, at least with the usual definition of continuity (though there is a sense in which such a function could be, for example, scott-continuous)
even in the case of a single discontinuity in the derivative, like in relu', you lose the intermediate value theorem and everything that follows from it; it's not an inconsequential or marginally relevant fact
Sure. But what part of this entirely worded in natural language, and very short statement made you think it was a technical, formal statement? I think you’re just taking an opportunity to flex your knowledge of basic calculus, and deliberately attributing intent to the author that isn’t there in order to look clever.
Regarding a function being discontinuous at a point outside its domain: if you take a completely naive view of what ‘discontinuous’ means, then I suppose you can say so. But discontinuity is just the logical negation of continuity. Observe:
To say that f: X —> Y (in this context, a real-valued function of real numbers) is continuous means precisely
∀x∈X ∀ε>0 ∃δ>0 |x - p| < δ ⇒ |f(x) - f(p)| < ε
and so its negation looks like
∃x∈X ⌐ …
that is, there is a point in X, the domain of f where continuity fails.
For example, you wouldn’t talk about a function defined on the integers being discontinuous at pi, would you? That would just be weird.
To prove the point further, observe that the set of discontinuities (according to your definition) of any given function would actually include every number… in fact every mathematical object in the universe — which would make it not even a set in ZFC. So it’s absurd.
Even more reasons to believe functions can only be discontinuous at points of their domain: a function is said to be discontinuous if it has at least one discontinuity. By your definition, every function is discontinuous.
…anyway, I said we were going to be petty. I’m trying to demonstrate this is a waste of time by wasting my own time.