Intermediate Value Theorem (Topology)
Let $f: X \to Y$ be a continuous mapping.
Let $a$ and $b$ are two points of $a, b \in X$ such that:
- $\map f a \prec \map f b$
- $r \in Y: \map f a \prec r \prec \map f b$
Then there exists a point $c$ of $X$ such that:
- $\map f c = r$
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 $B = f \sqbrk X \cap r^\succ$
$A$ and $B$ are disjoint by construction.
$A$ and $B$ are also non-empty since one contains $\map f a$ and the other contains $\map f b$.
Suppose there is no point $c$ such that $\map f c = r$.
- $f \sqbrk X = A \cup B$
so $A$ and $B$ constitute a separation of $X$.
Hence by Proof by Contradiction:
- $\exists c \in X: \map f c = r$
which is what was to be proved.
- 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.