Exploring the Non-Existence of a Maximum in the Set (0,2)

[ChiralSuperfields]

Homework Statement

I am trying to find the maximum of the set of real numbers in the open interval ##(0,2) ##

Relevant Equations

##(0,2)##

For this,

I am trying to understand why the set ##(0,2)## has no maximum. Is it because if we say for example claim that ##a_0 = 1.9999999999## is the max of the set, then we could come along and say that ##a_0 = 1.9999999999999999999999999999## is the max, can we continue doing that a infinite number of times so long that ##a_0 < 2##.

Many thanks!

 

[FactChecker]

I am trying to understand why the set ##(0,2)## has no maximum. Is it because if we say for example claim that ##a_0 = 1.9999999999## is the max of the set, then we could come along and say that ##a_0 = 1.9999999999999999999999999999## is the max, can we continue doing that a infinite number of times so long that ##a_0 < 2##.

Yes. Within (0,2), you can get as close as you want to 2, but 2 itself is not in the set (0,2). The definition specifies that the maximum must be in the set. So (0,2) has no maximum point.

 

[fresh_42]

I didn't understand your sentence after "max,".

Assume that ##(0,2)## has a maximum ##m<2.## Then ##m<m+\dfrac{2-m}{2}=\dfrac{2+m}{2}<2## which cannot be since ##m## was already the maximum. This is a contradiction, so there is no maximal number.

If you prefer the positive reasoning, then given any number ##m_0\in (0,2)## then ##m_1=\dfrac{2+m_0}{2}## is a number greater than ##m_0## and still smaller than ##2.## Now, we can proceed with that new number and define ##m_2= \dfrac{2+m_1}{2}.## This results in an infinite sequence
$$
0<m_0<m_1<m_2<\ldots < 2
$$
that gets closer and closer to ##2## but never ends. If this was what you wanted to say, then the answer is 'yes'.