Can you explain me what does it mean this property of the supremum and infinum?
If $A \subset \mathbb{R}$ and $c \in \mathbb{R}$, then we define $cA = \left\{ cx \: \middle| \: x \in A \right\}$.
If $c \geq 0$, then $\sup c A = c \sup A$, $\inf cA = c \inf A$.
If $c <0$ then $\sup cA = c \inf A$, $\inf c A = c \sup A$.
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The supremum or sup is the "least upper bound".
If A is a set and A is bounded above, then any M >= a for all a $\in$ A is an upper bound.
If x is the "least upper bound" than two things must be true:
1) x is an upper bound. It must be larger than all a in A.
2) for any y < x, y is not an upper bound. This means for any y < x there is some a $\in$ A such that y < a <= x.
A couple things to note:
i) sup A may or may not be in the set A. If it is in the set A then it is the largest element in the set. If it isn't in the set than the set does not have a single largest ellement.
ii) Depending on your metric space some bounded sets might not have a least upper bound. In Q for example, the set {$q \in Q | q^2 < 2$} is bounded above but there is no single rational number that is the smallest number that is as large or larger than every element in the set. (R however is not one of these spaces.)
iii) R, the real numbers, have the "least upper bound property" which means if A $\subset$ R and A is bounded above, the A will always have a least upper bound. As long as you are talking about subsets of R, then you can ignore ii).
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inf, meanwhile is the "greatest lower bound", it's the same concept but for sets that are bounded below.
One thing to note. If a space has the least upper bound property it also has the greatest lower bound property.
In R, if a set A is bounded above it has a sup. If B is bounded below it has an inf.
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So some examples.
Well, this is the simplest. Let A = {$x \in R | x < 27$}. A is bounded above because 100 > a for all a $\in$ A, and 100 is an upper bound. But 100 is obviously not the least upper bound, because 99, 57, 32 etc are all also upper bounds. Obviously the least upper bound is 27 but how does that fit the definition and can we prove it?
First of all 1) 27 is larger (or equal) than all a in A. This by definition of A. So 27 is an upper bound 2) if y < 27 then y < $\frac{y + 27}{2}$ < 27. So $\frac{y + 27}{2} \in$ A and y is not an upper bound. So 27 is the least upper bound. 27 = sup A.
A, meanwhile, is not bounded below so there is no inf.
Let B = {$x \in R| 2 \le x < 27$}. This is bounded above and bounded below. sup B = 27 by the exact same argument above. inf B = 2 by a similar argument. Note: sup B is not in B and B does not have a largest element. inf B is in B and B and inf B is the smallest element of B.
Think about what happens if $A$ is ain interval, for example $A=(1,2)$, when $\sup A=2$ and $\inf A=1$.
Then, $10 A = (10,20)$, and $\sup(10 A) = 20=10\cdot 2=10\sup A$ and $\inf (10A)=10=10\cdot 1=10\inf A$
On the other hand, $-5 A = (-10, -5)$, and $\sup(-5A)=-5=-5\inf A$, and $\inf (-5A)=-10=-5\sup A$