MATH 410.500
Spring 2014
Exam #2 Solutions
1. (a) true, (b) false, (c) false, (d) true, (e) true, (f) true, (g) false, (h) true, (i) false, (j)
false.
2. (b) Let U = {Ui }i∈I be a open cover of A ∪ B. Then U is an open cover of both A and
B, and
assumption there exist finite sets E, F ⊆ I such that
S so by the compactness
S
A ⊆ i∈ESUi and B ⊆ i∈F Ui . Then {Ui }i∈E∪F is a finite subcollection of U such that
A ∪ B ⊆ i∈E∪F Ui , which shows that A ∪ B is compact.
(c) We can write R2 as the union of the closed balls Br (0) centered at 0 with radii
ranging over r ∈ R. Since a closed ball is bounded, it is compact by the HeineBorel theorem. However, R2 is not compact. For example, {Br (0)}r∈R is an open
cover of R2 which does not have a finite subcover. Indeed if {Bri (0)}i=1,...,n is a finite
subcollection of this cover then its union would not contain the point (r, 0) where
r = 1 + max{r1 , . . . , rn }.
3. (b) It is enough to check
that each coordinate tends to 0 as (x, y) → (0, 0). Given
√
an ε > 0, set δ = √ε. Then for all (x, y) ∈ R2 satisfying k(x, y)k < δ we have
x2 + y 2 = k(x, y)k2 < δ = ε, and so x2 + y 2 → 0 as (x, y) → (0, 0). Next notice that
for all x, y ∈ R we have (x − y)2 ≥ 0 and (x + y)2 ≥ 0 and hence x2 + y 2 ≥ |2xy| by
expanding. Therefore when (x, y) 6= (0, 0) we have
2 2 x y x y
x2 + y 2 ≤ 2xy ≤ |x|.
Now given an ε > 0 set δ = ε. Then for all (x, y) 6= (0, 0) satisfying k(x, y)k < δ we
have
2 xy x2 + y 2 ≤ |x| ≤ k(x, y)k < ε,
showing that x2 y/(x2 + y 2 ) → 0 as (x, y) → (0, 0).
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