What is the difference between minimum and infimum?
I have a great confusion about this.
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2 Answers
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The minimum is attained, the infimum isn't necessarily.
Example.
Let $f(x) = \frac{1}{x}$. Then $f$ has no minimum value on the interval $(0,\infty)$. The minimum is the smallest element in the set. That is $$ \min\{f(x)\mid x\in (0,\infty)\} $$ doesn't exist because there is not smallest number in the set.
Another example is the minimum of the set $S = (0,1) = \{x\mid 0<x<1\}$. Here again there isn't a smallest number $$ \min\{x\mid 0<x<1\} $$ doesn't exist.
The infimum of a set $S$ is defined as the greatest number that is less than or equal to all elements of S (from Wikipedia). The infimum is also sometimes called the greatest lower bound.
It is a fact that every non empty set (bounded below) of real numbers has an infimum. But, as we saw, not every real set has a minimum.
So in the example $$ \inf\{f(x)\mid x\in (0,\infty)\} = 0. $$
Note that the infimum and the minimum can be the same. Consider for example $S = \{1,2,3,\dots\}$. Then the infimum and minimum is both $1$.
Consider this other example. If $f$ is a continuous function on a closed interval $[a,b]$, then it is a fact that $f$ attains a minimum over that interval. So here again $$ \inf\{f(x)\mid x\in [a,b]\} = \min\{f(x)\mid x\in [a,b]\}. $$
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14$\begingroup$ Is it correct to say that the infimum may not belong to the set in consideration, whereas the minimum must belong to it? $\endgroup$ Commented Mar 14, 2019 at 18:00
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2$\begingroup$ Yes that is correct @YanKingYin $\endgroup$– user496277Commented Feb 22, 2021 at 6:51
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2$\begingroup$ is it fair to say that distinguishing between inf and min only makes sense for infinite sets? $\endgroup$ Commented Apr 24, 2021 at 20:57
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2$\begingroup$ Yes, finite sets are compact, and compact sets contain their minimum. $\endgroup$ Commented Oct 26, 2023 at 17:41
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minimum is reached, infimum (may) not. That is, the numbers of the form $1/n$ have an inf (that is, 0), while the natural numbers have a min (that is, 1).
