Download Let E be the set of all p ∈ Q suc

2.12 Regard Q, the set of all rational numbers, as a metric space, with d(p, q) = |p − q|.
Let E be the set of all p ∈ Q such that 2 < p2 < 3. Show that E is closed and
bounded in Q, but that E is not compact. Is E open in Q?
Rudin’s Ex. 16
Proof Since for any p ∈ E, we have 1 < p, since otherwise 1 ≥ p2 , which contradicts
to the definition of E. Similarly, we have p < 2. Hence E is bounded.
To show that E is closed in Q, we assume√x ∈ Q is a limit point of E. Apparently,
x2 6= 2 and x2 6= 3. If x2 < 2, we let r1 = 2 − |x| > 0. If z ∈ Nr1 (x), then
|z| ≤ |x − z| + |x| < r1 + |x| =
√
2,
2
which implies that z < 2, or z ∈
/ E. This contradicts to the assumption that
x is a limit
point
of
E.
Thus,
necessarily
2 < x2 . Similarly, if x2 > 3, we let
√
r2 = |x| − 3 > 0. For any z ∈ Nr2 (x), since
√
|z| ≥ |x| − |x − z| > |x| − r2 = 3,
we have z ∈
/ E, which again contradicts to the assumption that x is a limit point of
E. This implies x2 < 3. Hence, 2 < x2 < 3. Therefore, we conclude that if x ∈ Q is
a limit point of E, then x ∈ E. This means that E is closed in Q.
To show that E is not compact, we take the collection {Un }∞
n=3 , where
Un = Q ∩ −(3 − n1 )1/2 , −(2 + n1 )1/2 ∪ (2 + n1 )1/2 , (3 − n1 )1/2 ,
n ≥ 3.
x ∈ Q since Q is the
“whole” space.
We should first argue
that if x2 = 3, then
x∈
/ Q.
√
√
2 and 3 are
well-defined by
Theorem 1.21.
Another way to show
E is closed is to prove
Ec = Q − E =
√
Q ∩√{(−∞,
√ − √3) ∪
(− 2, 2) ∪ ( 3, ∞)}.
E is bounded and
closed in Q, but not in
R. So, it is not
necessarily compact.
Each Un is open in Q by Theorem 2.30. For any x ∈ E ⊂ Q, since 2 < x2 < 3, there
is a positive integer n ≥ 3, such that
2+
1
n
< x2 < 3 − n1 ,
by the Archimedean property. Hence x ∈ Un . It follows that the collection {Un }∞
n=3
forms an open cover of E. Let {Un }n∈I , |I| < ∞, be any finite subcollection. Put
N = maxn∈I {n}. Then, since Q is dense in R by Theorem 1.20, there exists a
rational number r satisfying 3 − N1 < r2 < 3. From this construction,we know that
r ∈ E, but r ∈
/ ∪n∈I Un . Hence, any finite subcollection of {Un }∞
n=3 does not cover
E. We conclude that E is not compact.
Finally, we show that E is open in Q.√ Suppose x ∈ √
E ⊂ Q, with 2 < x2 < 3. Take
a number r such that 0 < r < min{ 3 − |x|, |x| − 2}. Consider a neighborhood
Nr (x) in Q:
Nr (x) = {z ∈ Q : |z − x| < r}.
For any z ∈ Nr (x), we have
√
3 − |x| + |x| = 3,
√
√
|z| ≥ |x| − |z − x| > |x| − (|x| − 2) = 2.
|z| ≤ |z − x| + |x| <
√
Hence Nr (x) ⊂ E. By the definition, E is open.
1
Or show E is open by
proving
√
√
E√= Q∩{(−
3, − 2)∪
√
( 2, 3)}.