517 Midterm 1
1. Suppose f : [a, b] → R is differentiable with f 0 (x) > 0 for all x ∈ (a, b). Prove that f is
strictly increasing.
Solution. Let x < y be in [a, b]. By MVT, for some c ∈ (x, y), f (y) − f (x) = f 0 (c)(y − x) > 0.
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2. Let f : X → R be continuous with X compact, and E = {x ∈ X : f (x) = 0}. Prove that E
is compact.
Solution. Note that {0} is closed in R. Since f is continuous, E = f −1 ({0}) is closed in X,
hence compact, since X is compact.
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3. Define f : R → R by f (x) = x2 . Prove that f is not uniformly continuous.
Solution. Suppose f is uniformly continuous. Then for = 1, there is δ > 0 such that
|x − y| < δ implies |f (x) − f (y)| < 1. Pick x = 1/δ and y = x + δ/2. Then |x − y| = δ/2 < δ
but |f (x) − f (y)| = |x − y||x + y| = (δ/2)(2/δ + δ/2) > 1, contradiction.
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4. Let {xn } be a sequence in a compact metric space X. Suppose all convergent subsequences
of {xn } have the same limit, x. Prove that {xn } converges to x.
Solution. Suppose {xn } does not converge to x. Then there is > 0 such that xn ∈
/ B (x)
for infinitely many n, so we can construct a subsequence {xnk } of {xn } whose range is disjoint
from B (x). But since X is compact, {xnk } has a converging subsequence, and its limit cannot
be x.
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5. Let X be a metric space with the Bolzano-Weierstrass property (every infinite subset of X
has a limit point in X). Without using any of our theorems about compactness, show that X is
totally bounded.
Solution. Let > 0. While possible, for n = 1, 2, ..., pick x1 , . . . , xn such that d(xi , xj ) ≥ for
1 ≤ i < j ≤ n. If this is possible for each n, we get an infinite subset {x1 , x2 , . . .} of X with no
limit point, contrary to the BW property. Thus, for some finite n we have min1≤i≤n d(xi , x) < for all x ∈ X, that is, X ⊂ B (x1 ) ∪ . . . ∪ B (xn ).
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