Let #X# be a subset of #\mathbb{R}# that is bounded above. This means that there is a real number #M# such that #x\le M# for each #x\in X#. Such a number #M# is also called an upper bound of #X#.
If #X# is a closed interval #\ivcc{a}{b}# that is bounded above, then #b# is the maximum of #X#, that is, the unique element of #X# with #x\le b# for all #x\in X#.
If #X# is an open interval #\ivoo{a}{b}# that is bounded above, then #b# is not the maximum of #X#, neither is any other number. Still, it is useful to have an interpretation of #b# in terms of #X#. The notion of supremum is very useful in this respect.
Let #X# be a nonempty set. We say that #s# is the supremum of #X# if it satisfies the following two conditions: #s# is an upper bound of #X# and #s# is the least upper bound of #X#, that is, for any #t# that is also an upper bound of #X#, we have #s\le t#.
Similarly, let #X# be a nonempty set. We say that a number #u# is the infimum of #X# if #-u# is a supremum of #-X#.
If #f# is a real function with domain #X#, then the supremum of #f# is the supremum of the image of #f# and similarly for the infimum, maximum, and minimum of #f#.
If #A# is a set or a function, we often write #\sup(A)# for the supremum and #\inf(A)# for the infimum of #A#. Likewise, we will write #\text{min}(A)# for the minimum of #A# and #\text{max}(A)# for the maximum of #A# if there is one.
For the open interval #\ivoo{a}{b}#, the infimum is #a# and the supremum is #b#, while the minimum and maximum do not exist.
For the closed interval \(\ivcc{a}{b}\) we have \[\begin{array}{rcl}\text{inf}(\ivcc{a}{b})&=&\text{min}(\ivcc{a}{b})= a\\ \text{sup}(\ivcc{a}{b})&=&\text{max}(\ivcc{a}{b})= b\end{array}\]
If both #s# and #t# are both suprema of #X#, then #s\le t# because #s# is a least upper bound and #t\le s# because #t# is a least upper bound, and so #s=t#. This shows that there is at most one supremum. The proof for the infimum is analogous.
Suppose that #X# is not empty, so it has an element #x_1#, and that it has an upper bound #s_1#. Then #\rv{x_1,s_1}# is a pair of numbers with #x_1\le s_1#. If #x_1=s_1#, then obviously #s_1# is the supremum of #X#. We want to show the supremum of #X# exists. Suppose that #s_1# is not the supremum. We proceed as follows for #n=1,2,\ldots#, starting with #n=1#. We set #m_{n} = \frac12(x_n+s_n)#. If #m_n# is an upper bound of #X#, we take #\rv{x_{n+1},s_{n+1}} = \rv{x_n,m_n}#. Otherwise, there is a point #u# in #X# with #u\gt m_n# and we take #\rv{x_{n+1},s_{n+1}} = \rv{u,s_n}#. This leads to a weakly descending sequence #s_1,s_2,\ldots# which is bounded below, so #s = \lim_{n\to\infty}s_n# exists. Now #x_1,x_2,\ldots# is a weakly ascending sequence, bounded above by #s#. Moreover, #\lim_{n\to\infty}x_n = s# because the distance between #s# and #x_{n+1}# is at most half the distance between #s# and #x_{n}#. If #s# is not the supremum of #X#, there would be a strictly smaller number #t# which is an upper bound of #X#. Because of #\lim_{n\to\infty}x_n = s#, the interval #\ivoc{t}{s}# would contain an element #x_N# for large enough #N#, contradicting that #t# is an upper bound. This shows that #s# is the least upper bound of #X#.
We conclude that the supremum of each nonempty set that is bounded above is unique. Similarly, the infimum of each nonempty set bounded below is unique.
If #X# is not empty and not bounded above, there is no upper bound and hence no supremum of #X#. We then write #\sup(X) = \infty#. Similarly, if #X# is not empty and not bounded below, the infimum of #X# does not exist and we write #\inf(X) = -\infty#.
The real function #f(x) = \frac{1}{x^2+1}# has maximum (and hence supremum) #1# (assumed at #x=0#) and infimum #0#, which is #\lim_{x\to\infty}f(x)#, but no minimum.
Instead of supremum, we also speak of least upper bound.
The infimum is also known as greatest lower bound.
We will now characterize these notions with a statement reminding us of limits.
Let #X# be a nonempty subset of the real numbers that is bounded above and let #s# be an upper bound of #X#. The number #s# is the supremum of #X# if and only if, for each positive number #\varepsilon# there exists an element #x# of #X# such that #s-\varepsilon \lt x#.
Let #X# be a nonempty subset of the real numbers that is bounded below and let #s# be a lower bound of #X#. The number #s# is the infimum of #X# if and only if, for each positive number #\varepsilon# there exists an element #x# of #X# such that #s+\varepsilon \gt x#.
Let #X# and #s# be as stated in the hypothesis. The proof for the case of an infimum is very similar to the case of a supremum. Therefore, we only restrict to the case of a supremum. Suppose that #s# is an upper bound of #X#.
#\Rightarrow# Assume that #s# is a supremum of #X# and let #\varepsilon# be an arbitrary positive number. Then #s-\varepsilon\lt s#, so #s-\varepsilon# is not an upper bound of #X#. By the definition of supremum for #s#, there exists an element #x# in #X# such that #s-\varepsilon\lt x#.
#\Leftarrow# Suppose next that, for each positive number #\varepsilon#, there exists an element #x# in #X# such that #s-\varepsilon\lt x#. Let #t# be an upper bound of #X#. Assume that #t\lt s#. Put #\varepsilon =s-t#. Then #\varepsilon\gt0#, so there exists an element #x# in #X# such that #t=s-\varepsilon \lt x#. This contradicts the condition that #t# is an upper bound of #X#. We conclude that #t\ge s#. This proves that #s# is the least upper bound of #X#, so #s # is a supremum of #X#.
In terms of limits: An upper bound #s# of a subset #X# of #\mathbb{R}# is a supremum of #X# if and only if there is a sequence #x_1,x_2,\ldots# of elements #x_i# of #X# such that #\lim_{n\to\infty}x_n = s#.
If #X# is the set of elements of a sequence #x_1,x_2,\ldots# of real numbers bounded above by #M#, then #\lim_{n\to\infty}x_n# need not exist, but #s = \sup(X)# is well defined and satisfies #s\le M#.
If, for instance #x_n = (-1)^n\cdot \frac{n-1}{n}#, then \[\sup\{x_n\mid n=1,2,\ldots\} = 1\phantom{xxx}\text{ and }\phantom{xxx}\inf\{x_n\mid n=1,2,\ldots\} = -1\]
If #x_1,x_2,\ldots# is a sequence with \[\sup\{x_n\mid n=1,2,\ldots\} =\inf\{x_n\mid n=1,2,\ldots\} \]then #\lim_{n\to\infty}x_n# does exist, and coincides with the supremum and the infimum of the set of all #x_n#.
If #a_i# #(i=1,2,\ldots)# is an increasing sequence, then its limit, if it exists, is equal to the supremum of the set #\{a_i\mid i=1,2,\ldots\}#.
If #a_i# #(i=1,2,\ldots)# is an decreasing sequence, then its limit, if it exists, is equal to the infimum of the set #\{a_i\mid i=1,2,\ldots\}#.
Let #f# be the function on # \ivoo{-\infty}{\infty} # given by
\[ f(x) = x-\left \lfloor x \right \rfloor\]
where # \lfloor x\rfloor # is the greatest integer smaller than or equal to #x#. Determine whether the maximum, the supremum, the minimum, and the infimum of #f# exist and, if so, determine its value.
Enter #\text{none}# if the value does not exist (so do not use #\infty#).
- #\max(f)=# # none#
- #\sup(f)= # # 1#
- #\min(f)=# # 0#
- #\inf(f)=# # 0#
The four values are the maximum, supremum, minimum, and infimum of the image of #f#. Therefore, we first determine this image.
The values of #f(x)# lie between #0# and #1#. The value #1# will never be attained because #x = \lfloor x \rfloor+1# would imply that #x# is an integer, so #\lfloor x \rfloor+1=x = \lfloor x \rfloor#, a contradiction. The value #0# is attained for each integer #x#.
Therefore, the image of #f# coincides with the interval #\ivco{0}{1}#. We conclude that #\max(f) = none#, #\sup(f) = 1#, #\min(f) = 0# and #\inf(f) =0#.
Infimum and supremum
