Continuous Real-Valued Function is not necessarily Bounded

Theorem

Let $S$ be a set.

Let $f: S \to \R$ be a continuous real-valued function.

Then $f$ is not necessarily bounded.

Proof

Let $S$ denote the (open) real interval $\openint 0 1$.

Let $f: S \to \R$ denote the reciprocal function:

$\forall x \in S: \map f x = \dfrac 1 x$

From Reciprocal Function is Continuous on Real Numbers without Zero, $f$ is continuous on $S$.

From Reciprocal Function is Unbounded on Open Unit Interval, $f$ is unbounded on $S$.

Hence the result, by Proof by Counterexample.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.1$: Motivation: Example $5.1.2$