Exercises: Chapter 1, Section 3

  1. Prove that ch1c1.png and ch1c2.png , show that ch1c3.png if and only if ch1c4.png for each ch1c5.png .

    Suppose that ch1c6.png for each i. Let ch1c7.png . Choose for each ch1c8.png , a positive ch1c9.png such that for every ch1c10.png with ch1c11.png , one has ch1c12.png . Let ch1c13.png . Then, if ch1c14.png satisfies ch1c15.png , then ch1c16.png . So, ch1c17.png .

    Conversely, suppose that ch1c18.png , ch1c19.png , and ch1c20.png is chosen as in the definition of ch1c21.png . Then, for each i, if ch1c22.png is in ch1c23.png and satisfies ch1c24.png , then ch1c25.png . So ch1c26.png .

  2. Prove that ch1c27.png is continuous at ch1c28.png if and only if each ch1c29.png is.

    This is an immediate consequence of Problem 1-23 and the definition of continuity.

  3. Prove that a linear transformation ch1c30.png is continuous.

    By Problem 1-10, there is an ch1c31.png such that ch1c32.png for all ch1c33.png . Let ch1c34.png and ch1c35.png . Let ch1c36.png . If ch1c37.png satisfies ch1c38.png , then ch1c39.png . So T is continuous at ch1c40.png .

  4. Let ch1c41.png .
    1. Show that every straight line through ch1c42.png contains an interval around ch1c43.png which is in ch1c44.png .

      Let the line be ch1c45.png . If ch1c46.png , then the whole line is disjoint from ch1c47.png . On the other hand, if ch1c48.png , then the line intersects the graph of ch1c49.png at ch1c50.png and ch1c51.png and nowhere else. Let ch1c52.png . Then ch1c53.png is continuous and ch1c54.png . Since the only roots of ch1c55.png are at 0 and ch1c56.png , it follows by the intermediate value theorem that ch1c57.png for all ch1c58.png with ch1c59.png . In particular, the line ch1c60.png cannot intersect ch1c61.png anywhere to the left of ch1c62.png .

    2. Define ch1c63.png by ch1c64.png if ch1c65.png and ch1c66.png if ch1c67.png . For ch1c68.png define ch1c69.png by ch1c70.png . Show that each ch1c71.png is continuous at 0, but ch1c72.png is not continuous at ch1c73.png .

      For each ch1c74.png , ch1c75.png is identically zero in a neighborhood of zero by part (a). So, every ch1c76.png is clearly continuous at 0. On the other hand, ch1c77.png cannot be continuous at ch1c78.png because every open rectangle containing ch1c79.png contains points of ch1c80.png and for all those points ch1c81.png , one has ch1c82.png .

  5. Prove that ch1c83.png is open by considering the function ch1c84.png with ch1c85.png .

    The function ch1c86.png is continuous. In fact, let ch1c87.png and ch1c88.png . Let ch1c89.png . If ch1c90.png , then by Problem 1-4, one has: ch1c91.png . This proves that ch1c92.png is continuous.

    Since ch1c93.png , it follows that ch1c94.png is open by Theorem 1-8.

  6. If ch1c95.png is not closed, show that there is a continuous function ch1c96.png which is unbounded.

    As suggested, choose ch1c97.png to be a boundary point of ch1c98.png which is not in ch1c99.png , and let ch1c100.png . Clearly, this is unbounded. To show it is continuous at ch1c101.png , let ch1c102.png and choose ch1c103.png . Then for any ch1c104.png with ch1c105.png , one has ch1c106.png . So,

    ch1c107.png

    where we have used Problem 1-4 in the simplification. This shows that ch1c108.png is continuous at ch1c109.png .

  7. If ch1c110.png is compact, prove that every continuous function ch1c111.png takes on a maximum and a minimum value.

    By Theorem 1-9, ch1c112.png is compact, and hence is closed and bounded. Let ch1c113.png (resp. ch1c114.png ) be the greatest lower bound (respectively least upper bound) of ch1c115.png . Then ch1c116.png and ch1c117.png are boundary points of ch1c118.png , and hence are in ch1c119.png since it is closed. Clearly these are the minimum and maximum values of ch1c120.png , and they are taken on since they are in ch1c121.png .

  8. Let ch1c122.png be an increasing function. If ch1c123.png are distinct, show that ch1c124.png .

    One has ch1c125.png . The function on the right is an increasing function of ch1c126.png ; in particular, ch1c127.png is bounded above by the quantity on the right for any ch1c128.png . Now assume that the ch1c129.png have been re-ordered so that they are in increasing order; let ch1c130.png . Now add up all the inequalities with this value of ch1c131.png ; it is an upper bound for the sum of the ch1c132.png and the right hand side ``telescopes" and is bounded above by the difference of the two end terms which in turn is bounded above by ch1c133.png .