1. Complete the following sentence "A set $U$ fails to be a neighborhood of a number $x$ when for every number $\delta >0,$ ..."
    A set $U$ fails to be a neighborhood of a number $x$ when for every number $\delta >0$ the interval MATH contains at least one number that does not belong to $U$.

  2. Explain carefully why the assertion
    MATH
    that was made above is true. You will need to make use of an earlier result.
    Since the number $0$ belongs to the interval MATH for every positive integer $n$ we have
    MATH
    Now we need to show that $0$ is the only number in the set MATH. Suppose that $x\neq 0$. Using the fact that MATH and that earlier result, choose a positive integer $n$ such that
    MATH

    MATH
    Since we have found a value of $n$ for which $x$ fails to belong to the interval MATH we know that
    MATH
    Thus $0$ is the only number in the set MATH, which is what we wanted to prove.

  3. Given that $x$ is an interior point of $U$ and that $U\subseteq V,$ explain why $x$ must be an interior point of $V.$
    Using the fact that $x$ is an interior point of $U$, choose $\delta >0$ such that
    MATH
    Since $U\subseteq V$ we therefore have
    MATH
    Therefore $x$ is an interior point of $V$.

  4. Suppose that $x$ is a real number and that MATH Prove that the following two conditions are equivalent:

    1. The set $U$ is a neighborhood of the number $x.$

    2. It is possible to find two numbers $a$ and $b$ such that
      MATH

    Solution: To prove that condition a implies condition b we assume that $U$ is a neighborhood of the number $x$. Choose $\delta >0$ such that
    MATH
    By defining $a=x-\delta $ and $b=x+\delta $ we obtain two numbers $a$ and $b$ such that $a<b$ and such that
    MATH

    Now to prove that condition b implies condition a we assume that condition b holds. Choose numbers $a$ and $b$ such that $a<b$ and such that
    MATH
    Using the fact that the interval $\left( a,b\right) $ is a neighborhood of $x$, choose $\delta >0$ such that
    MATH
    Since MATH we have shown that $U$ is a neighborhood of $x$.

  5. Suppose that $x$ and $y$ are two different real numbers. Prove that it is possible to find a neighborhood $U$ of $x$ and a neighborhood $V$ of $y$ such that MATH
    Solution: We may assume, without loss of generality, that $x<y$. Choose a number $c$ between $x$ and $y$. The intervals MATH and MATH are, respectively, neighborhoods of $x$ and $y$ and the intersection of these two intervals is empty.

  6. Given that $S$ is a set of real numbers and that $x$ is an upper bound of $S,$ explain why $S$ cannot be a neighborhood of $x$.
    Solution: If $\delta $ is any positive number then, since all of the numbers between $x$ and $x+\delta $ must lie in $R\setminus S$, the interval MATH cannot be included in $S$.

  7. Given that a set $S$ of real numbers is nonempty and bounded above, explain why neither $S$ nor MATH can be a neighborhood of $\sup S.$
    Solution: We see from Exercise 6 that $S$ is not a neighborhood of $\sup S$. Now we observe that, whenever $\delta >0$, since the number $x-\delta $ fails to be an upper bound of $S$, there must be members of $S$ in the interval MATH. Therefore, whenever $\delta >0$, the interval MATH fails to be included in the set $R\setminus S$ and so $R\setminus S$ must also fail to be a neighborhood of $x$.

  8. Suppose that $A$ and $B$ are sets of real numbers and that $x$ is an interior point of the set $A\cup B.$ Is it true that $x$ must either be an interior point of $A$ or an interior point of $B?$
    Hint: The answer is no. Give an example to show what can go wrong.

  9. Suppose that $A$ and $B$ are sets of real numbers and that $x$ is an interior point both of $A$ and of $B.$ Is it true that $x$ must be an interior point of the set $A\cap B?$
    Yes it is true. Suppose that $x$ is an interior point of both of the sets $A$ and $B$. Choose a number $\delta _{1}>0$ such that
    MATH
    and choose a number $\delta _{2}>0$ such that
    MATH
    We now define $\delta $ to be the smaller of the two numbers $\delta _{1}$ and $\delta _{2}$ and we observe that
    MATH

  10. Suppose that $x$ and $y$ are real numbers and that $U$ is a neighborhood of $y.$ Prove that the set $V$ defined by
    MATH
    is a neighborhood of the number $x+y.$
    Solution: We need to find a number $\delta >0$ such that the interval MATH is included in $V$. Using the fact that $U$ is a neighborhood of $y$, we choose $\delta >0$ such that
    MATH
    Now given any number $t$ in the interval MATH we deduce from the fact that
    MATH
    that
    MATH
    and therefore we know that $t-x\in U$. Therefore, since
    MATH
    we know that $t\in V$ and we have shown that
    MATH

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