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The nonempty real set $T$ is bounded below. Its infimum $m=\inf T$ satisfies $m\le t$ for every $t\in T$, and every other lower bound is at most $m$. It is the greatest lower bound.
Equivalently, for every $\varepsilon>0$ some $t\in T$ satisfies $m\le t<m+\varepsilon$. Values can approach arbitrarily closely from above; $m$ need not belong to $T$. Completeness of the real numbers ensures an infimum exists for a nonempty set bounded below.
For a function, $\inf_{x\in S}f(x)$ applies this definition to its range. For $f(x)=x$ on $(0,1)$, the infimum is 0. Zero bounds all values below, while every positive candidate fails: some smaller positive value remains in the interval.
If the infimum is attained at an admissible input, it is a minimum. Zero is excluded from this open interval, so no minimum exists. Without a lower bound, no finite real infimum exists; writing $-\infty$ uses an extended-real convention.
The approximation condition identifies the greatest lower bound. If a larger lower bound $b$ existed, the choice $\varepsilon=b-m$ applies. The definition would then produce an element $t<b$, contradicting the lower-bound property of $b$. Merely showing that every element exceeds the candidate is insufficient: every smaller number would pass that weaker test as well.
Definitions and notation: Basic Analysis I.
Related concepts: A function’s minimum: the smallest attained value.
In plain language
An infimum resembles the highest floor that still fits below every value. A lower floor is possible, but any upward shift violates the condition for some value. For $(0,1)$ the floor sits at zero even though no number stands on it. An unoccupied floor does not imply a shared positive gap; values may approach the boundary arbitrarily closely.
Example
The constructed set is $(0,1)$. The example tests lower-bound candidates and approaches zero.
Infimum: the greatest lower bound
A loose lower bound
Minus one is below all values, but zero gives a higher lower bound.
Step 1 of 4: A loose lower bound
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A loose lower bound
Minus one is below all values, but zero gives a higher lower bound.
- Candidates
- −10current
An invalid higher bound
Candidate 0.1 fails because admissible value 0.05 is smaller.
- Candidate; counterexample
- 0.1excluded0.05included
Approaching from above
Positive values 0.1, 0.01, and 0.001 remain in the set and approach zero.
- Admissible values
- 0.10.01current0.001included
The bound is unattained
Zero is the exact lower bound but is excluded from the set.
- Infimum; membership
- 0included0 ∉ (0,1)excluded
For any positive candidate $b$, an interval element lies below it: the choice $t=\min(b/2,1/2)$ gives a number that is positive, below 1, and below $b$. One construction of a counterexample thus replaces an endless inspection of decimal samples. Using the same construction for accuracy $\varepsilon$ gives an element between zero and $\varepsilon$. None of these steps declares zero an admissible element of the interval.
An infimum must be a lower bound that cannot be raised. Attainment is a separate check.