Exercises on Limit Points

  1. Prove that MATH
    Given any number $x$, the interval MATH can contain at most two integers. We know that a neighborhood of a limit point of a set must always contain infinitely many members of that set and so we conclude that no number $x$ can be a limit point of the set $Z$ of integers.

  2. Prove that MATH
    Suppose that $x$ is any real number. To show that $x$ is a limit point of $Q$, suppose that $\delta >0$. Since there are rational numbers in the interval MATH we know that the set MATH.

  3. Prove that MATH
    If $x$ is any negative number then the interval MATH is a neighborhood of $x$ that fails to contain any members of the set MATH. Thus a negative number can not be a limit point of MATH.
    If $x$ is any positive number then the interval MATH is a neighborhood of $x$ which fails to contain infinitely many members of the set MATH. To see why, note that if $n$ is a positive integer then the conditionMATHcan hold only if $n<\frac{2}{x}$. Therefore no positive number can be a limit point of MATH.
    Finally, we need to explain why $0$ must be a limit point of MATH. Suppose that $\delta >0$. Choose an integer MATH and observe thatMATHfrom which we deduce that the set MATH must be nonempty.

    1. Give an example of an infinite set that has no limit point.
      As we saw in Exercise 1, the infinite set $Z$ has no limit point.

    2. Give an example of a bounded set that has no limit point.
      A finite set like $\left\{ 2\right\} $ will not have any limit points. We could also look at the empty set $\emptyset $.

    3. Give an example of an unbounded set that has no limit point.
      As we saw in Exercise 1, the infinite set $Z$ has no limit point.

    4. Give an example of an unbounded set that has exactly one limit point.
      The unbounded set MATH has only the limit point $0$.

    5. Give an example of an unbounded set that has exactly two limit points.
      The setMATHhas the two limit points $0$ and $1$. We can see this directly or we can use the assertion proved in Exercise 6 below.

  5. Prove that if $A$ and $B$ are sets of real numbers and if $A\subseteq B$ then MATH
    Suppose that $A$ and $B$ are sets of real numbers and that $A\subseteq B$. Suppose that $x$ is a limit point of $A$. We need to explain why $x$ has to be a limit point of $B$. Suppose that $\delta >0$. Since the set MATH is nonempty and sinceMATHwe deduce that the set MATH is nonempty.

  6. Prove that if $A$ and $B$ are sets of real numbers then MATHSolution: Since $A\subseteq A\cup B$ we know that MATH and similarly we know that MATH. ThusMATHNow suppose that a number $x$ fails to belong to the set MATH. Choose a number $\delta _{1}>0$ such that the interval MATH contains only finitely many members of the set $A$. Choose a number $\delta _{2}>0$ such that the interval MATH contains only finitely many members of the set $B$. We now define $\delta $ to be the smaller of the two numbers $\delta _{1}$ and $\delta _{2}$ and we observe that, although $\delta >0$, the interval MATH contains only finitely many members of the set $A\cup B$. Therefore no number that lies outside the set MATH can be a limit point of $A\cup B$ and we conclude thatMATH

  7. Is it true that if $A$ and $B$ are sets of real numbers then MATHWhat if $A$ and $B$ are closed? What if $A$ and $B$ are open? What if $A$ and $B$ are intervals?
    The answers are no, no, no and no. Look at the following example:MATHThese two sets are closed andMATHwhileMATHNow look at the following example:MATHIn this caseMATHandMATH

  8. Is it true that if $\overline{D}=R$ then MATH?
    Hint: Yes.
    Suppose that $\overline{D}=R$. We know that whenever $a$ and $b$ are real numbers and $a<b$ there must be members of $D$ lying between $a$ and $b$. Now suppose that $x$ is a real number. To show that $x$ is a limit point of $D$, suppose that $\delta >0$. Since there must be members of $D$ in the interval MATH we conclude that the set MATH is nonempty.

  9. Given that a set $S$ of real numbers is nonempty and bounded above but that $S$ does not have a largest member, prove that $\sup S$ must be a limit point of $S$. State and prove a similar result about $\inf S.$
    To show that $\sup S$ is a limit point of $S$, suppose that $\delta >0$. Since MATH and since $\sup S$ is the least upper bound of $S$ the number $\sup S-\delta $ fails to be an upper bound of $S$. Choose a member $x$ of $S$ such that $\sup S-\delta <x$. Since $x\leq \sup S$ and since $\sup S$ does not belong to $S$ we have $x<\sup S$. We conclude thatMATH

  10. Given any set $S$ of real numbers, prove that the set $L\left( S\right) $ must be closed.
    Solution: We shall show that any number that fails to belong to $L\left( S\right) $ must fail to belong to MATH. Suppose that MATH. Choose a number $\delta >0$ such that the interval MATH contains only finitely many members of $S$. Given any number $t$ in the interval MATH, it follows from the fact that MATH is a neighborhood of $t$ and the fact that MATH contains only finitely many members of $S$ that $t$ is not a limit point of $S$. ThusMATHand we have shown, as promised, that $x$ does not belong to MATH.

  11. Prove that if a set $U$ is open then MATH
    Of course MATH. Now suppose that $x\in \overline{U}$. To show that MATH, suppose that $\delta >0$. Using the fact that $x\in \overline{U}$, choose a number $y$ in the set MATH. Using the fact that the set MATH is open, choose $\varepsilon >0$ such thatMATHWe have now found more than one member of $U$ that belongs to the interval MATH and so we know thatMATH

  12. Suppose that $S$ is a set of real numbers, that MATH and that $\delta >0.$ Prove that there exist two different numbers $x$ and $y$ in $S$ such that MATH
    Solution: Choose a limit point $t$ of the set $S$. Using the fact that the interval MATH contains infinitely many members of $S$, choose two different members $x$ and $y$ of $S$ that lie in the interval MATH. We observe that MATH.