Is this definition of boundedness correct? Hi I'm trying to learn precalc and one thing I'm stumped on is the concept of boundedness. From what I understand boundedness is whether or not a function has an absolute maximum and minimum. If it has only max then it is bounded above, otherwise if it has only a minimum it is bounded below. If it is both then its bounded, if it has neither it is not bounded. Is this definition correct? I'm trying to use this logic to check if $y=\sqrt{13-x^2}$ is bounded in any way or not.
 A: You have the right idea, but a function bounded above need not have an absolute maximum, a function bounded below need not have an absolute minimum, and a bounded function need not have either. A real-valued function on $\Bbb R$ is bounded above if there is some real number $m_a$ such that $f(x)\le m_a$ for all $x\in\Bbb R$, and it is bounded below if there is some real number $m_b$ such that $f(x)\ge m_b$ for all $x\in\Bbb R$. It is bounded if it is bounded both above and below. There is no requirement that $f(x)=m_a$ for some $x$, even if $m_a$ is the smallest real number $m$ with the property that $f(x)\le m$ for all $x\in\Bbb R$, and a similar statement is true of $m_b$.
An example of a bounded function that does not have an absolute maximum or minimum is the function $f(x)=\tan^{-1}x$: $-\frac{\pi}2<f(x)<\frac{\pi}2$ for all $x\in\Bbb R$, but $f(x)$ has neither an absolute maximum nor an absolute minimum.
A: More simply, the "maximum" of a set of numbers is the largest number in the set and the "minimum" is the smallest number in the set.
The "open interval", (0, 1), the set of all numbers x, 0< x< 1, has 1, and every number larger than 1, as "upper bound" and 1 is the "least upper bound" but is NOT the "maximum" because 1 is not itself in the set.  Similarly, any number 0 or less  is a  "lower bound".  0 is the "greatest lower bound" but is not the "minimum" because it is not itself in the set.
A: Boundedness does not implies that the function has an absolute maximum or an absolute minimum.
Let consider for example $f(x)=\frac1x$ for $x>0$ which is bounded below but there is not a minimum (there exists an infimum that is zero).
What is true is that when an absolute maximum [or minimum] exists then $f(x)$ is bounded above [or below].
