Homework 3 Key
1. (1 point) The rationals are countable and dense in the reals, as shown
in class!
2. (1 points) Again, the rationals are a fine example. And a bounded
example is the set of all rationals in [0, 1] (since this set is not closed).
3. (3 points) The set E is not countable, for if we map 2 ↔ 0 and 6 ↔ 1
we obtain a bijection from E to the set of all 0 − 1 sequences, shown
to be uncountable.
The set E is not dense in [0, 1]. For example, the interval (0.34, 0.35)
contains no elements of E, so no number in (0.34, 0.35) can be a limit
point for E.
The set E is bounded. Is it closed? Yes! To see this, let p be a limit
point for E. I claim p cannot have a decimal expansion whose first
digit is anything except 2 or 6. To see this, suppose in contradiction
that, for example, that p = 0.d · · · with d = 0, 1. Then 0 < p < 0.2
and some small neighborhood of p contains only numbers with decimal
expansions that start with a digit of 1 or 0. A similar argument works
for d = 3, 4, 5, 7, 8, 9. Thus if p is a limit point for E, p must have
leading digit 2 or 6. A similar argument works for the second digit—if
p is a limit point, the second digit must be a 2 or 6, and so on. Any
limit point for E must have a decimal expansion consisting entirely of
2’s or 6’s, and so E is closed. Since E is bounded, E is compact.
4. (3 points) There are many ways to proceed, but to start note that
2
n +n
1
|2n − 5|
−
2n2 + 5 2 = 2(2n2 + 5)
(no absolute values on the denominator, since it’s always positive.)
Now here’s the scratch work part of the proof: First, clearly
|2n − 5|
2n
n
<
=
.
2(2n2 + 5)
2(2n2 + 5)
2n2 + 5
And if we note that 2n2 < 2n2 + 5 we see that
n
1
n
< 2 =
.
+5
2n
2n
2n2
1
1
If we obtain 2n
< ϵ, this will imply what we want, and this happens if
we take n > 1/(2ϵ). So here’s the formal proof.
Formal Proof: Let N = ⌈ 2ϵ1 ⌉+1. Then for n ≥ N we have n > 1/(2ϵ),
or 1/(2n) < ϵ, and so
2
n + n 1
|2n − 5|
2n + 5 − 2 = 2(2n2 + 5)
n
<
2
2n + 5
1
<
2n
< ϵ.
This completes the proof.
2