# Is this proof of the Intermediate Value Theorem sufficient?

This proof looks sound to me, but it seems too simple to work.

Proof of intermediate value theorem:

Suppose $$t \in [m,M]$$, with $$m$$ and $$M$$ the minimum and maximum values of $$f$$ on $$[a,b]$$, respectively, and $$f(c) \neq t$$ for any $$c\in [a,b]$$. By the extreme value theorem, $$m \leq f(c) \leq M$$ $$\forall c \in [a,b]$$. Therefore $$f(c) \in [m,M]$$ $$\forall c \in [a,b].$$ Since $$t \in [m,M]$$, there is a $$c \in [a,b]$$ such that $$f(c) = t$$.

• it is completely wrong. What justifies the last sentence? – mookid Nov 2 '14 at 21:46
• I was with you until the last sentence. – vadim123 Nov 2 '14 at 21:46
• No, this is not a proof. The same argument would give us that there is an integer between $2$ and $3$. You still have not said anything that would not apply to that (false) case. For instance, nowhere have you mentioned (or used) continuity of $f$, which I assume is one of your assumptions. – Andrés E. Caicedo Nov 2 '14 at 21:47

The proof is not sound. From "$f(c)\in[m,M]$ for all $c\in[a,b]$" and $t\in[m,M]$ you cannot infer "There exists $c\in[a,b]$ such that $f(c)=t$". For example
• $n^2 \in \mathbb N_0$ for all $n\in\mathbb N_0$
• $2\in \mathbb N_0$
• therefore(?) $n^2=2$ for some $n\in\mathbb N_0$