Intermediate Value Theorem (Topology)

Theorem
Let $X$ be a connected topological space.

Let $\struct {Y, \preceq, \tau}$ be a totally ordered set equipped with the order topology.

Let $f: X \to Y$ be a continuous mapping.

Let $a$ and $b$ are two points of $a, b \in X$ such that:
 * $\map f a \prec \map f b$

Let:
 * $r \in Y: \map f a \prec r \prec \map f b$

Then there exists a point $c$ of $X$ such that:
 * $\map f c = r$

Proof
Let $a, b \in X$, and let $r \in Y$ lie between $\map f a$ and $\map f b$.

Define the sets:
 * $A = f \sqbrk X \cap r^\prec$ and $B = f \sqbrk X \cap r^\succ$

where $r^\prec$ and $r^\succ$ denote the strict lower closure and strict upper closure respectively of $r$ in $Y$.

$A$ and $B$ are disjoint by construction.

$A$ and $B$ are also non-empty since one contains $\map f a$ and the other contains $\map f b$.

$A$ and $B$ are also both open by definition as the intersection of open sets.

Suppose there is no point $c$ such that $\map f c = r$.

Then:
 * $f \sqbrk X = A \cup B$

so $A$ and $B$ constitute a separation of $X$.

But this contradicts the fact that Continuous Image of Connected Space is Connected.

Hence by Proof by Contradiction:
 * $\exists c \in X: \map f c = r$

which is what was to be proved.

Also see

 * Intermediate Value Theorem of calculus, which follows as a corollary from this by considering $\R$ under the order topology and noting that Subset of Real Numbers is Interval iff Connected.