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$L : (0, \infty) \to R$ is a continuous strictly increasing bijection

Statement

The function

$L : (0, \infty) \to R$

is a continuous strictly increasing bijection.

Facts & Assumptions

Given: $L$ on $(0, \infty)$ and a target $r \in R$ .

[L1]

$L$ is continuous and strictly increasing on $(0, \infty)$ (The integral logarithm is continuous and strictly increasing on $(0, \infty)$ ).

[L2]

For every real $r$ , there are positive $a, b$ with $L (a) < r < L (b)$ (The integral logarithm is unbounded above and below).

[L3]

A continuous real function on $[a, b]$ takes every value between its endpoint values (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ).

Proof

technique · direct

1.1

Strict increase in [L1] makes $L$ injective.

1.2

By [L2], choose positive $a, b$ with $L (a) < r < L (b)$ . Strict increase in [L1] then implies $a < b$ .

L2L1

2.1

The restriction of $L$ to $[a, b]$ is continuous by [L1], so [L3] gives $x \in (a, b)$ with $L (x) = r$ . Thus $L$ is surjective onto $R$ .

step 1.2L1L3

3.1

Steps 1.1 and 2.1 show that $L$ is a bijection, and continuity and strict increase are already supplied by [L1].

step 1.1step 2.1L1∎

Depends on

Used by

The integral exponential E:ℝ→(0,∞) as the inverse of L Definition

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 84 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

OpenStax, Calculus Volume 1, Section 6.7 (standard reference, not scraped)

L:(0,∞)→RL:(0,∞)→R is a continuous strictly increasing bijection