Proving Existence of Zero with Least Upper Bounds

In summary, to show that there is a largest x in [a,b] with f(x) = 0, we use the least upper bound property to show that the set of points where f(x) = 0 has a supremum. Assuming this supremum is not in the set, we can find a contradiction and prove that it must be in the set.
  • #1
andilus
8
0

Homework Statement



sup problem
if f is continuous on [a,b] with f(a)<0<f(b), show that there is a largest x in [a,b] with f(x)=0

Homework Equations

i think it can be done by least upper bounds, but i dun know wat is the exact prove.

The Attempt at a Solution

 
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  • #2
Start by showing that there is one x in [a,b] such that f(x) = 0. Then form the set
[tex]S = \{x \in [a,b] | f(x) = 0\}[/tex]
You have already shown that S is non-empty and you know that it's bounded, so it must have a supremum. Let
[tex]x_0 = \sup\,S[/tex]
Since [itex]x_0[/itex] is an upper bound for S, if we can show [itex]x_0 \in S[/itex] we have shown that it's the largest element in S. So all you need to do is show [itex]x_0 \in S[/itex]. The easiest way to do this is to assume [itex]x_0 \notin S[/itex], i.e. [itex]y_0 = f(x_0) \not= 0[/itex]. Now since f is continuous at [itex]x_0[/itex] we can find some [itex]\delta > 0[/itex] such that if [itex]x \in (x_0-\delta,x_0 + \delta)[/itex] then [itex]f(x) \in (0,2y_0)[/itex] (let [itex]\epsilon = |y_0|[/itex]). Now [itex]x_0 - \delta[/itex] is an upper-bound for S, but less than [itex]x_0[/itex].
 
  • #3
but how to prove x0 is in [a,b]?
 
  • #4
i seem to know...
if x0<a, obviously wrong.
if x0>b, we can find some x such that b-[tex]\delta[/tex]<x<b satisfying that x is upper bound of the set.
 

What is the definition of "least upper bound"?

A least upper bound (or supremum) is the smallest number that is greater than or equal to all the numbers in a set. In other words, it is the smallest possible upper limit of a set of numbers.

Why is proving the existence of zero with least upper bounds important?

Proving the existence of zero with least upper bounds is important because it is a fundamental concept in mathematics and helps establish the properties of the number system. It also serves as a building block for more complex concepts in mathematics.

How is the existence of zero with least upper bounds proven?

The existence of zero with least upper bounds is proven using the completeness axiom, which states that every non-empty set of real numbers that is bounded above has a least upper bound. This axiom is accepted as a fundamental truth and cannot be proven.

What are some applications of the existence of zero with least upper bounds?

The existence of zero with least upper bounds has many practical applications, such as in calculus, where it is used to prove the convergence of sequences and series. It is also used in optimization problems, where the least upper bound represents the optimal solution.

Are there any alternative ways to prove the existence of zero with least upper bounds?

Yes, there are alternative ways to prove the existence of zero with least upper bounds, such as using the axiom of Archimedes or the nested intervals theorem. However, these methods are equivalent to using the completeness axiom and do not provide a different proof of the existence of zero with least upper bounds.

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