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Theorem 18.1

Let $f $ be a continuous real-valued function on a closed interval $[a, b] \implies f$ is bounded.

Moreover, $\exists x_0, y_0 \in [a, b], \forall x \in [a, b], f(x_0) \leq f(x) \leq f(y_0)$.

Theorem 18.2: Intermediate Value Theorem

If $f $ is a continuous real-valued function on an interval $I $, then $f $ has the intermediate value property on $I $:

$a, b \in I$, $a < b$. $\forall y $ lies between $f(a)$ and $f(b) $, $\exists x \in (a, b)$ such that $f(x) = y $.

Proof [expand]

We assume $f(a) < y < f(b) $; the other case is similar.

Let $S = \set{ x \in [a, b] \mid f(x) < y }$. Since $a $ belongs to belongs to $S$.

$\therefore S \neq \emptyset $.

$x_0 = \supr{ S } \in [a, b]$.

$\forall n \in \N $, $x_0 - \frac{ 1 }{ n } $ is not an upper bound for $S $, so there exists $s_n \in S $ s.t. $x_0 - \frac{ 1 }{ n } < s_n \leq x_0 $.

Thus $\lim s_n = x_0 $

$f(s_n) < y \forall n$,

$\therefore f(x_0) = \lim f(s_n) \leq y$

Let $t_n = \min{ \set{ b, x_0 + \frac{ 1 }{ n }}} $. Since $x_0 \leq t_n \leq x_0 + \frac{ 1 }{ n } $ we have $\lim t_n = x_0 $.

Each $t_n $ belongs to $[a, b] $ but not to $S $, so $f(t_n) \geq y $ for all $n $.

$\therefore f(x_0) = \lim f(t_n) \geq y$

$f(x_0) = y$.

Corollary 18.3

If $f$ is a continuous real-valued function on an interval $I $, then the set $f(I) = \set{ f(x) \mid x \in I } $ is also an interval or a single point.

Theorem 18.4

Let $f $ be a continuous strictly increasing function on some interval $I$.

Then $f(I) $ is an interval $J $ bt the previous corollary and $\inv{ f } $ represents a function with domain $J$. The function $\inv{ f } $ is a continuous strictly increasing function on $J $.

Theorem 18.5

Let $g$ be a strictly increasing function on an interval $J$, $g(J)$ is an interval $I$ $\implies g$ is continuous on $J $.

Theorem 18.6

$f $ is one-to-one and continuous function on an interval $I \implies f $ is strictly increasing or strictly decreasing.