Reciprocal Function is Unbounded on Open Unit Interval
Jump to navigation
Jump to search
Theorem
Let $A = \openint 0 1$ denote the open unit interval.
Let $f: A \to \R$ be the reciprocal function:
- $\forall x \in A: \map f x := \dfrac 1 x$
Then $f$ is unbounded.
Proof
Let $K \in \R_{>0}$.
Then:
- $\exists x \in \R: 0 < x < \dfrac 1 K$ such that $x < 1$.
Then we have:
- $\map f x = \dfrac 1 x > K$
So whatever $K$ may be, it can never be large enough to be a bound of $f$ on $\openint 0 1$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.1$: Motivation: Example $5.1.2$