Continuous Real Function Bounded on Finite Subdivision

Theorem

Let $A = \closedint a b$ be a closed real interval of the set $\R$ of real numbers.

Let $f: A \to \R$ be a continuous real function on $A$.

Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ be a finite subdivision of $A$ such that:

each $\closedint {x_j} {x_{j + 1} }$ is a neighborhood of some $a_j$ such that $f$ is bounded on $\closedint {x_j} {x_{j + 1} }$.

Then $f$ is bounded on $A$.

Proof

Follows directly from:

Continuous Real Function is Bounded on Neighborhood of Argument

and:

Mapping is Bounded on Union iff Bounded on Each Component/Real-Valued Function.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.1$: Motivation: Step $4$