I was given this problem the following problem in my exam:

A function continuous on $[a,b]$ attains a minimum value on $[a,b]$.

Note: proof should not involve compact sets or sequences.

Help me with this proof please. Thank you.


1 Answer 1


Using the completeness axiom in Real numbers: For any closed interval in $\mathbb R $ ,there exists a greatest lower bound,
called the infimum,which is equal to the minimum of that set.

Since the function is continuous and the domain is a closed interval, it follows that its range is also a closed interval,So the completeness axiom holds for the Range of that function, i.e. We get a minimum.

See also : https://en.m.wikipedia.org/wiki/Extreme_value_theorem

ADD:To see why the range would be a closed set, Consider $c$ to be a point in the interval $[a,b]$, and assume it is one of the end points of the function's range. Then if the range was not a closed interval,Then $f(c)$ would not be in the range,and thus :$$ \lim_{x\to c} f(x) = f(c) $$ would not hold, Since the function would not be defined on the point $c$, and we would get a contradiction that the function is continuous on the interval $[a,b]$,Therefore it is a closed set.

  • 2
    $\begingroup$ That is a proof? $\endgroup$
    – socrates
    Nov 28, 2016 at 22:42
  • 1
    $\begingroup$ @socrates leave behind the socratic method.. ;P $\endgroup$
    – user378947
    Nov 28, 2016 at 22:46
  • $\begingroup$ Why wouldn't it be?. It completely follows from the completeness axiom and the defintion of continuity. Although if the function was also said to be differetiable ,you could use the fact the it had to be monotone in a sub interval $[m,n]and then show that a monotone (strictly increasing or decreasing function) attains its extremum at it's end points,m,n and to say that the minumum exists... but that is an overkill and it does follow from the completeness axiom. $\endgroup$ Nov 28, 2016 at 22:48
  • 3
    $\begingroup$ How do you prove that the range is closed without using sequences or passing through compactness? $\endgroup$
    – Del
    Nov 28, 2016 at 22:56
  • $\begingroup$ By using the condition that the function is continuous.see:en.m.wikipedia.org/wiki/Extreme_value_theorem $\endgroup$ Nov 28, 2016 at 23:05

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