Exercises on Supremum And Infimum

  1. Suppose that $A$ is a nonempty bounded set of real numbers that has no largest member and that $a\in A.$ Prove that MATH

    You saw in an earlier exercise that the sets $A$ and MATH have exactly the same upper bounds.

  2. Given that $A$ and $B$ are sets of numbers, that $A$ is nonempty, that $B$ is bounded above and that $A\subseteq B,$ explain why $\sup A$ and $\sup B$ exist and why $\sup A\leq \sup B.$
    Since $A$ is nonempty and $A\subseteq B$ we know that $B$ is nonempty. Since $A\subseteq B$ we know that any upper bound of $B$ must be an upper bound of $A$. Therefore, since $B$ is bounded above we know that that $A$ is bounded above. Finally, since $\sup B$, being an upper bound of $B$, must be an upper bound of $A$, and since $\sup A$ is the least upper bound of $A$ we have $\sup A\leq \sup B$.

  3. Given that $A$ is a nonempty bounded set of numbers, explain why $\inf A\leq \sup A.$
    Using the fact that $A$ is nonempty we choose a member $a$ of $A$. We know that
    MATH

  4. It is given that $A$ and $B$ are nonempty bounded sets of real numbers, that for every $x\in A$ there exists $y\in B$ such that $x<y$ and for every $y\in B$ there exists $x\in A$ such that $y<x.$ Prove that $\sup A=\sup B.$

    Solution: We need to show that the two sets $A$ and $B$ have exactly the same upper bounds and for this purpose we shall show that a number fails to be an upper bound of $A$ if and only if it fails to be an upper bound of $B$.

    Suppose that $u$ fails to be an upper bound of $A$. Choose a member $x$ of $A$ such that $u<x$. Using the given property of $A$ and $B$ we now choose a member $y$ of the set $B$ such that $x<y$ and, since $u<x<y$, we conclude that $u$ can't be an upper bound of $B$. We can show similarly that a number that fails to be an upper bound of $B$ must also fail to be an upper bound of $A$.

  5. Suppose that $A$ and $B$ are nonempty sets of real numbers and that for every $x\in A$ and every $y\in B$ we have $x<y.$ Prove that $\sup A\leq \inf B.$ Give an example of sets $A$ and $B$ satisfying these conditions for which $\sup A=\inf B.$
    Given any member $y$ of the set $B$ it follows from the fact that $x<y$ for every $x\in A$ that $y$ must be an upper bound of $A$. In other words, every member of $B$ is an upper bound of $A$ and the fact that $B$ is nonempty shows that $A$ is bounded above. A similar argument shows that every member of $A$ is a lower bound of $B$ and the fact that $A$ is nonempty guarantees that $B$ is bounded below. Thus $\sup A$ and $\inf B$ exist.
    Given any member $y$ of $B$, the fact that $y$ is an upper bound of $A$ and $\sup A$ is the least upper bound of $A$ tells us that $\sup A\leq y$. In other words, $\sup A$ is a lower bound of $B$. Since $\inf B$ is the greatest lower bound of $B$ we deduce that $\sup A\leq \inf B$.

  6. Suppose that $A$ and $B$ are nonempty sets of real numbers and that $\sup A=\inf B.$ Prove that for every number $\delta >0$ it is possible to find a member $x$ of $A$ and a member $y$ of $B$ such that $x+\delta >y.$

    Solution: Suppose that $\delta >0$. Using the fact that
    MATH
    and that $\sup A+\delta $ is therefore not a lower bound of $B$, choose a member $y$ of the set $B$ such that
    MATH
    From the fact that
    MATH
    we deduce that $y-\delta $ is not an upper bound of $A$ and, using this fact, we choose a member $x$ of $A$ such that $y-\delta <x$. In this way we have found $x\in A$ and $Y\in B$ such that $x+\delta >y$.
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  7. Suppose that $A$ and $B$ are nonempty sets of real numbers, that $\sup A\leq \inf B$ and that for every number $\delta >0$ it is possible to find a member $x$ of $A$ and a member $y$ of $B$ such that $x+\delta >y.$ Prove that $\sup A=\inf B.$

    Solution: To obtain a contradiction, assume that $\sup A<\inf B$ and define
    MATH
    We observe that $\delta >0$. Now for all $x\in A$ and $y\in B$, it follows from the facts that $x\leq \sup A$ and $\inf B\leq y$ that
    MATH
    which tells us that $x+\delta \leq y$. Therefore it is impossible to find $x\in A$ and $y\in B$ such that $x+\delta >y$ and we have reached the desired contradiction.

  8. Suppose that $A$ is a nonempty bounded set of real numbers, that $A$ has no largest member and that $x<\sup A.$ Prove that there are at least two different members of $A$ lying between $x$ and $\sup A.$
    Since $x$ is less than the least upper bound of $A$ we know that $x$ can't be an upper bound of $A$. Using the fact that $x$ isn't an upper bound of $A$ we choose a member $a$ of $A$ such that $x<a$. Since $A$ has no largest member it must have a member larger than the member $a$. We choose a member $b$ of $A$ such that $a<b$. Since $A$ has members greater than $b$ we deduce that
    MATH
    and we have found two different members of $A$ between $x$ and $\sup A$.

  9. Suppose that $A$ is a nonempty bounded set of real numbers, that $\delta >0$ and that for any two different members $x$ and $y$ of $A$ we have MATH Prove that $A$ has a largest member. You can find a hint to the solution of this exercise in a forthcoming theorem.

  10. Suppose that $S$ is a nonempty bounded set of real numbers, that $\alpha =\inf S$ and $\beta =\sup S$, and that every number that lies between two members of $S$ must also belong to $S$. Prove that $S$ must be one of the four intervals MATH MATH MATH MATH See a coming theorem for a solution of this exercise.

  11. Suppose that $A$ is a nonempty bounded set of real numbers, that $\alpha =\inf A$ and that $\beta =\sup A.$ Suppose that
    MATH
    Prove that MATH You will find a solution to this exercise in the next section.

  12. Suppose that $A$ is a set of numbers and that $A$ is nonempty and bounded above. Suppose that $q$ is a given number and that the set $C$ is defined as follows:
    MATH
    Prove that the set $C$ is nonempty and bounded above and that
    MATH

    This exercise is a special case of the next exercise because MATH.

  13. Suppose that $A$ and $B$ are nonempty bounded sets of numbers and that the sets $A+B$ and $A-B$ are defined as above. Prove that
    MATH
    and
    MATH
    Solution: We shall prove that
    MATH
    Step 1: We want to show that the number $\sup A+\sup B$ is an upper bound of the set $A+B$. Suppose that $x\in A+B$. Using the definition of $A+B$ we choose a number $a\in A$ and a number $b\in B$ such that $x=a+b$. Since $a\leq \sup A$ and $b\leq \sup B$ we see that
    MATH

    Step 2: We want to show that the number $\sup A+\sup B$ is actually the least upper bound of the set $A+B$. Suppose that $u$ is any upper bound of the set $A+B$. Given any number $a\in A$ and $b\in B$ we have $a+b\leq u$. Therefore, whenever $b\in B$ we know that the inequality
    MATH
    holds for all $a\in A$. Therefore, whenever $b\in B$, the number $u-b$ is an upper bound of $A$ and must satisfy the condition
    MATH
    which we can also write as
    MATH
    We conclude that $u-\sup A$ is an upper bound of $B$ and so
    MATH
    which we can write as
    MATH
    Thus $\sup A+\sup B$ is the least upper bound of $A+B$, as promised.

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