| Math 425-1 | Exam 2 Comments | Fall 2008 |
1. To prove that a continuous function on a sequentially compact set
has a bounded range:
Comment. The proof of this is almost identical with the proof
of Lemma 3.10 on page 60, but this is simpler, since in 3.10 one uses the
fact that a closed bounded interval is sequentially compact, whereas here
the domain is given to be sequentially compact.
2. To investigate the continuity of the function defined to be 0 on
the irrationals and the identity on the rationals.
Comment. This is similar to Example 3.3 on page 55, except
that a simple epsilon-delta argument shows that the given function here is
continous at 0.
3. To prove the uniform continuity of a given function h.
Comment. A simple algebraic argument shows that
| h(u) -
h(v)| ≤
| u -
v | for the
given function, and then the proof of uniform continuity is the same as
that in Exercise 11, Sect. 3.2, on page 69 (for Lipschitz continuous
functions).
4. To prove that for a bounded continuous function
f on the real numbers, the equation f(x) =
x has a solution.
Comment. This is similar to Exercise 5, Sect. 3.3, on page 65.
| What has been is what will be, and what has been done is what will be done; there is nothing new under the sun. Is there a thing of which it is said, "See, this is new"? It has already been, in the ages before us. (Ecclesiastes 1: 9-10) |
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