Minimum vs lowerbound What is the difference between the minimum value and the  lower bound of a function? To me, it seems that they are the same.
 A: Let $A$ be a non-empty set of real numbers, such as the set of all negative reals; or the interval $[0, 1]$, a.k.a. the set of $x$ with $0 \leq x \leq 1$; or the set of positive reals.


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*We say $A$ is bounded below if there is some number $L$ such that $L \leq a$ for all $a$ in $A$. Any such $L$ is a lower bound of $A$.

*We say a number $\underline{a}$ in $A$ is a minimum or a smallest element of $A$ if $\underline{a} \leq a$ for all $a$ in $A$.

*We say a number $I$ is a greatest lower bound or an infimum of $A$ if (i) $I$ is a lower bound of $A$, and (ii) no number larger than $I$ is a lower bound of $A$.
These concepts are closely related, but logically distinct:


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*If $A$ is bounded below, there are always infinitely many lower bounds.

*A minimum element is unique (if it exists at all): If $\underline{a}$ and $\underline{a}'$ are minima of $A$, then $\underline{a} \leq \underline{a}'$ because $\underline{a}$ is a minimum and $\underline{a}'$ is an element of $A$. Reversing the roles shows $\underline{a}' \leq \underline{a}$.) A minimum is a lower bound of $A$ that happens to be an element of $A$.

*An infimum is unique by a similar argument for minima. Conceptually, an infimum is "just like a minimum", except it need not be an element of $A$.
The set of negative reals is not bounded below.
The interval $[0, 1]$ is bounded below, and every real number $L \leq 0$ is a lower bound; the minimum exists, and is $L = 0$.
The set of positive real numbers is bounded below, and every real number $L \leq 0$ is a lower bound; there is no minimum, but $I = 0$ is the infimum.
Each term applies to real-valued functions if we let $A$ be the image, a.k.a. the set of output values of $f$. Particularly, if a function $f$ achieves a minimum $L$, then $L$ is a lower bound for $f$. Again, the minimum value of $f$ (if it exists at all) is unique, while a lower bound for $f$ is not unique.
The function $f(x) = -1/x^{2}$ (defined for $x \neq 0$) achieves every negative real value, and is therefore not bounded below.
The function $f(x) = \cos^{2} x$ achieves precisely values in $[0, 1]$. Every real number $L \leq 0$ is a lower bound of this $f$, and $0$ is the minimum value (achieved at odd integer multiples of $\pi/2$).
The function $f(x) = 1/x^{2}$ (defined for $x \neq 0$) is bounded below, and has infimum $0$, but has no minimum.
A: Simple example: consider $f(x)=2^{-x}$.  For any $x\in\Bbb R$, $f(x)>0$.  So $0$ is a lower bound for $f$. But $f$ does not have a minimum.  For any $x>0$, no matter how small, there is a positive integer $n$ so that $2^{-n}<x$.
