Intermediate Value Theorem

Theorem
Let $$f:I\to \mathbb{R}$$ be a real function, defined on any interval $$I\subset \mathbb{R},$$ and consider two points $$a,b\in I.$$ Given $$f$$ continuous at $$I.$$

If $$k\in\mathbb R$$ satisfies the relation $$f(a)k\}.$$

These subsets are not empty. (Since $$a$$ belongs to one of them and $$b$$ belongs to the other one.) They constitute a partition of $$[a,b]$$ in two subsets, since, supposed that $$f(x)\ne k$$ for all $$x\in\,[a,b],$$ every element of $$[a,b]$$ belongs to some of $$I\,'$$ or $$I\,'',$$ and obviously it can't belong both.

So, exists a point $$c\in\,[a,b]$$ which belongs to one of the subsets $$I\,'$$ or $$I\,,$$ and which is the limit of a certain sequence $$x_n$$ of points of the other subset; $$c=\lim_{n\to\infty}x_n;$$ suppose, for example that $$c\in I'$$ and that $$x_n\in I$$ for $$n\in\mathbb N.$$ (In the same fashion one can consider that $$x'_n\in I'$$ and that $$c\in I'',$$ the argument is the same.) Since $$c\in\,[a,b]\subset I,$$ the function $$f$$ is continuous at $$c,$$ and given $$c=\lim_{n\to\infty}x_n,$$ it has to verify that $$f(c)=\underset{n\to \infty }{\mathop{\lim }}\,f\left( x_{n} \right),$$ which is impossible since $$f(c)k$$ for $$n\in \mathbb{N}.$$

Given $$\epsilon =k-f(c)>0$$ of $$f(c)$$ there's no one of the $$f\left( x_{n} \right).\,\to \!\!|\!\!\leftarrow$$

This contradiction proves the Theorem.

Q.E.D.