Question

problem1&2 thx！

interval in R is a set IC R such that for all <y < z in R, if E I and z e I then Recall that an points yE I. We call an interval non-degenerate if it contains at least two (1) Let I be a nondegenerate interval in R, and suppose f: IR is continuous (a) Show that f[] is an interval in R. (b) Show that if I is closed and bounded, then so is f[I]. (c) Show by example that if I is not closed then f[I] need not be bounded or closed (2) Show that "a composition of uniformly continuous functions is uniformly continuous." That is, prove that if f: A R and g BR are uniformly continuous, where ran(f) C B then gof: AR is uniformly continuous

Answer 1

(1)

(a) take any points x and y in I

f(x) and f(y)

E f(I)

for f(I) to be an interval, every d such that
F(x<d< f(y)
should be in
f(I)

Intermediate Value Theorem. Let f be a function which is continuous on the closed interval [a, b]. Suppose that d is a real number between f(a) and f(b); then there exists c in [a, b] such that f(c) = d

Hence is an interval

f(I)

(b)

Extreme Value Theorem. Suppose that f is a function which is continuous on the closed interval [a, b]. Then there exist real numbers c and d in [a, b] such that

• f has a maximum value at x = c and
• f has a minimum value at x = d.

f(c) and f(d)

E f(I)

By extreme value theorem if I is a closed and bounded interval f(I) is bounded and closed

(c) Let I = (0,1)

f

Here f(I) is neither closed nor bounded

(2)

We know that f(x) is uniformly continuous and g(x) is uniformly
continuous.
Therefore, given
any h > 0 there exists d_0 > 0 such that
|f(x) - f(y)| < h whenever |x - y| < d_0.
Also, there exists d_1 > 0 such that if |s - t| < d_1
then |g(s) - g(t)| < d_0. Let d=min(d_0,d_1).
Let s=f(x) and t=f(y).
So, if |x - y|< d then |f(x) - f(y)|<d_0
and |g(f(x)) - g(f(y))|< h. Thus the composition of 2 uniformly
continuous functions is also uniformly continuous.