Math 342
Homework 5
Daniel Walton
January 16, 2014
Section 11.1 : 4, 6, 10
Exercise (4). Suppose f : Rn → R is continuous and that f (u) > 0
if u ∈ R has at least one rational component. Prove f (u) ≥ 0 for
all points u ∈ R.
Solution (4). For every u ∈ Rn with a rational component, we
already have that f (u) ≥ 0. So, we need only be concerned with
elements with all irrational components. Let u = (u1 , . . . , un ) ∈ Rn .
Fix > 0. The continuity of f implies that there exists δ > 0 such
that if dist(u, v) < δ, dist(f (u), f (v)) < . Since the rationals are
dense in R, we can find a rational number v1 such that u1 < v1 <
u1 + δ. Then 0 < v1 − u1 < δ. We create the vector v ∈ Rn identical to u except with v1 in the first
p
p component so that dist(u,v)=
2
(u1 − v1 ) + . . . + (un − vn ) = (u1 − v1 )2 + 0 + . . . + 0 = |u1 −
v1 | < δ. By continuity and f (v) > 0, we have that
|f (u) − f (v)| < f (u) − f (v) > −
f (u) > f (v) − > 0 − = −.
Since > 0 can be chosen arbitrarily small, we have that f (u) ≥ 0.
Exercise (6). Suppose that the function f, g : Rn → R are both
continuous. Prove that the set A = {u ∈ Rn |f (u) = g(u) = 0} is
closed in Rn .
Solution (6). Let {uk }k∈N be a sequence contained in A converging
to u ∈ Rn . We show that u ∈ A. Let > 0. By continuity
of f and g, there exists δ1 , δ2 such that ||uk − u|| < δ1 implies
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|f (uk )−f (u)| < , and ||uk −u|| < δ2 implies |g(uk )−g(u)| < . Let
δ = min(δ1 , δ2 ). Since uk → u, ∃K ∈ N such that k ≥ K guarantees
||uk − u|| < δ. Then by continuity of f and g, |f (uk ) − f (u)| < and |g(uk ) − g(u)| < . Since uk ∈ A, we have f (uk ) = g(uk ) = 0
∀k. Then we have |f (u)|, |g(u)| < . Since was arbitrary, f (u) =
g(u) = 0, which completes the proof.
Alternatively, we can utilize Corollary 11.13 as follows: Notice that
A = {u ∈ Rn |g(u) ≤ 0} ∩ {u ∈ Rn |g(u) ≥ 0} ∩ {u ∈ Rn |f (u) ≤
0} ∩ {u ∈ Rn |f (u) ≥ 0}. Since f and g are continuous, these four
sets satisfy the hypotheses of Corollary 11.13 and therefore must be
closed. By Proposition 10.17, the intersection of these four sets is
also closed and hence A is closed.
Exercise (10). Let O be and open subset of Rn and suppose that
the function f : O → R is continuous. Suppose that u is a point in
O at which f (u) > 0. Prove that there is an open ball B about u
such that f (v) > f (u)/2 for all v ∈ B.
Solution (10). Choose = f (v)/2. By the continuity of f , we
know that there exists δ1 > 0 such that ||u − v|| < δ1 implies
|f (u) − f (v)| < . Since O is open, there exists an open ball of
radius r about u contained in O. Choose δ = min(δ1 , r). We are
guaranteed to have Bδ ⊆ O since Bδ ⊆ Br . Then by continuity of
f , if v ∈ Bδ , |f (u) − f (v)| < = f (u)/2, which implies f (v) >
f (u) − f (u)/2 = f (v)/2. This concludes the proof.
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