If I've read Rudin correctly the definitions for limits and for continuity are
∀ϵ>0 ∃δ>0. 0 < d(x, p) < δ ⇒ d(f(x), q) < ϵ
∀ϵ>0 ∃δ>0. d(x, p) < δ ⇒ d(f(x), f(p)) < ϵ
I don't understand the significance of the missing "0 < " bit in the definition of continuity. The definition of continuity is equivalent to:
∀ϵ>0 ∃δ>0 both statements are true:
x ≠ p ∧ d(x, p) < δ ⇒ d(f(x), f(p)) < ϵ (1)
x = p ∧ d(x, p) < δ ⇒ d(f(x), f(p)) < ϵ (2)
and statement (1) is equivalent to the following statements
(1) ⇔ 0 < d(x, p) ∧ d(x, p) < δ ⇒ d(f(x), f(p)) < ϵ
⇔ 0 < d(x, p) < δ ⇒ d(f(x), f(p)) < ϵ
and statement (2) is equivalent to the following statements:
(2) ⇔ x = p ∧ 0 < δ ⇒ 0 < ϵ
⇔ x = p ∧ ⊤ ⇒ ⊤
⇔ x = p ⇒ ⊤
⇔ ⊤
We find then that the definition of continuity is equivalent to
∀ϵ>0 ∃δ>0 both statements are true:
0 < d(x, p) < δ ⇒ d(f(x), f(p)) < ϵ
⊤
which is of course equivalent to
∀ϵ>0 ∃δ>0. 0 < d(x, p) < δ ⇒ d(f(x), p(q)) < ϵ (⋆)
which is equivalent to
f(x) → f(p) as x → p
And so the definition Rudin gives is equivalent to
f(x) is continuous at p iff f(x)→f(p) as x → p
Am I missing something?
One thought; is theorem 4.6 there simply because of this wierd behaviour of limits; If p is an isolated point then f(x) → L as x → p for every possible value of L! And so if you use (⋆) as the definition of continuity and aren't careful you might end up wrongly stating that:
Let p be an isolated point, we know that f(x) → L as x → p, and also that L ≠ f(p), therefore f(x) is not continuous at p.
But that error stems not from any problem with (⋆), rather it stems from us wrongly thinking that a function can have only one limit at a point when in these weird cases it can have multiple.