Math 351 (Real Analysis) Solutions to the Sixth Homework Set. 1 ...

1. Let E = {r ∈ Q ≥0 | r 2 < 7}.
Math 351 (Real Analysis)
Solutions to the Sixth Homework Set.
(a) Determine whether E is an open set as well as a closed set. Explain.
Solution: E is not open, since any ɛ neighbourhood of a point in E contains an
irrational point (which doesn’t belong to E). Likewise, E is not closed, because
R − E is not open: Any ɛ neighbourhood of a point in R − E between 0 and √ 7
contains an rational point (which doesn’t belong to R − E).
(b) Find the limit points of E. Justify your answer.
Solution: Since {1/n} ⊆ E converges to 0, 0 is a limit point of E. Likewise,
any point x ∈ (0, √ 7] is a limit point of E, by using a sequence of rational points
from E that converges to x (courtesy of the density of Q applied to increasingly
small neighbourhoods around x). In summary, the set of limit points of E is
[0, √ 7].
2. The following sets are not compact. For each set, provide a sequence (without proof)
contained in the given set that does not possess a subsequence converging to a limit
in the set.
(a) Q ∩ [0, 1]
Solution: Choose a sequence of rational numbers in [0, 1] that converges to an
irrational number in [0, 1].
(b) R
Solution: Choose an unbounded sequence of real numbers, such as {n}.
(c) { 1 n | n ∈ N}
Solution: Note that {1/n} converges to 0 /∈ { 1 n | n ∈ N}.
3. Use the δ − ɛ definition of a limit to establish each of the following are true.
(a) lim(3x − 5x + 1) = 3
x→2
Proof: Given ɛ > 0, let δ = min{1, ɛ/10}. Then, whenever |x − 2| < δ,
|(3x 2 − 5x + 1) − 3| = |3(x − 2) 2 + 7(x − 2)|
≤ 3|x − 2| 2 + 7|x − 2|
< 10|x − 2|
< 10(ɛ/10)
= ɛ.
√
(b) lim x = 3
x→9
Proof: Given ɛ > 0, let δ = 3ɛ. Then, whenever |x − 9| < δ,
| √ x − 3| =
|x − 9|
√ x + 3
≤
|x − 9|
0 + 3
< 3ɛ/3 = ɛ.

4. Give δ − ɛ definitions for the one-sided limits lim f(x) = L and lim f(x) = M.
x→c− x→c +
Solution: We say that lim f(x) = L if given any ɛ > 0, there exists δ > 0 such that
x→c− whenever c − δ < x, |f(x) − L| < ɛ.
Likewise, lim f(x) = M if given any ɛ > 0, there exists δ > 0 such that whenever
x→c +
x < c + δ, |f(x) − M| < ɛ.
5. Let g : A → R and assume that f is a bounded function on A ⊆ R (i.e., there exists
M > 0 satisfying |f(x)| ≤ M for all x ∈ A). Show that if lim g(x) = 0, then
x→c
lim g(x)f(x) = 0 as well.
x→c
Solution: By hypothesis on f, there exists M > 0 satisfying |f(x)| ≤ M for all
x ∈ A. Moreover, given ɛ > 0, there exists δ > 0 such that |x − c| < δ implies
|g(x) − 0| < ɛ/M.
Thus, |x − c| < δ implies that |g(x)f(x) − 0| < (ɛ/M)M = ɛ, as required.