Intermediate Value Theorem

maths ✒ ivt

Suppose $ f : [a,b] \to \mathbb{R} $, f continuous over $ [a,b] $, $ c \in (f(a), f(b)) $.

Then $ \exists d \in (a,b) ~ $ such that $ ~ f(d) = c $.

Proof

This one is less hassle than the usual interval bisection one!

Let $ S = \{x \in [a,b] ~ : ~ f(x) \le c \} $.

Then S non-empty since $ a \in S $ and S bounded above by b.

Therefore $ s = sup ~ S $ exists.

Suppose $ f(s) > c $. Then $ f(s) - c > 0 $ so by the continuity of f,
$ \exists \delta $ s.t. $ \forall 0 < | x - s | < \delta, ~ | f(x) - f(s) | < f(s) - c $.

Therefore $ f(x) > c $ on $ (s - \delta, s + \delta) $.

But this implies $ s - \delta $ is an upper bound of S which contradicts $ sup S = s $.

Suppose instead that $ f(s) < c $. Then by a similar argument, $ f(x) < c $ on $ (s - \delta', s + \delta') $ for some $ \delta' > 0 $.

Therefore $ s + \frac \delta 2 \in S $ but $ s + \frac \delta 2 > s = sup ~ S $ so we have a contradiction.

By trichotomy we conclude that $ f(s) = c $. So write $ d = s $.

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