Introduction to Analysis – MA 511 – Fall 2011 – R. Pollack
HW #8 solutions
Rudin Chapter 3 24(c,d,e)
24c) We must take a sequence of equivalence classes of Cauchy sequences, say {Pn }, and produce a limit of
this sequence. That is Pn ∈ X ∗ for each n, and let {pn,j }j denote a representative Cauchy sequence from
Pn . We must produce a Cauchy sequence {qn }n such that if Q is the corresponding equivalence class of
{qn }n , then {Pn } → Q in X ∗ .
We first build {qn }n . Since {pn,j }j is Cauchy, for ε = n1 , there is some Mn such that r, s ≥ Mn implies
d(pn,r , pn,s ) < n1 . We define
qn = pn,Mn .
We must check that {qn }n is in fact a Cauchy sequence. To this end, fix ε > 0 and pick N1 large enough
so that 1/N1 < ε/3. Also, since {Pn } is Cauchy in X ∗ , there is some N2 such that r, s > N2 implies
∆(Pr , Ps ) < ε/4. Thus, for r, s > N2 , we have limn d(pr,n , ps,n ) < ε/4. In particular, there is some N3 such
that n > N3 and r, s > N1 implies d(pr,n , ps,n ) < ε/3 (since ε/3 > ε/4).
Now fix r, s ≥ N , and set N = max{N1 , N2 }. Then for any M > max{Mr , Ms , N3 }, we have
d(qr , qs ) = d(pr,Mr , ps,Ms )
≤ d(pr,Mr , pr,M ) + d(pr,M , ps,M ) + d(ps,M , ps,Ms )
< 1/r + ε/3 + 1/s
< ε/3 + ε/3 + ε/3 = ε.
Hence, {qn } is Cauchy.
Now, let Q denote the equivalence class of {qn } and we must prove that {Pn } → Q in X ∗ . To this end,
fix ε > 0 and pick N large enough so that 1/N < ε/3 and so that ∆(Pr , Ps ) < ε/4 for r, s > N . Then for
n > N , we need to show that ∆(Pn , Q) ≤ ε. Since ∆(Pn , Q) = limi d(pn,i , qi ), we consider
d(pn,i , qi ) = d(pn,i , pi,Mi ) ≤ d(pn,i , pn,M ) + d(pn,M , pi,M ) + d(pi,M , pi,Mi ).
As before, as long as i and M are greater than Mn , the first term will be less than ε/3. By choosing M
greater than Mi , the last term will be less than ε/3. Lastly, if i > N , by choosing M large enough, we can
make the middle term as close to ε/4 as we want – say less than ε/3. Thus, for i large enough, d(pn,i , qi ) < ε
which implies ∆(Pn , Q) ≤ ε as desired.
24d) We have ∆(Pp , Pq ) = limn d(p, q) = d(p, q) as desired.
24e) We need to check that ϕ(X) is dense in X ∗ . So take a Cauchy sequence {pn } representing some element
of X ∗ and fix some ε > 0. We need to find an element in ϕ(X) within ε of {pn }. By the definition of Cauchy,
there is some N such that m, n ≥ N implies d(pn , pm ) < ε/2. Consider the constant Cauchy sequence
{pN } ∈ ϕ(X). Then
∆({pN }, {pn }) = lim d(pN , pn ) ≤ ε/2 < ε
n
∗
as desired. Thus, ϕ(X) is dense in X .
Lastly, we need to check that if X is complete, then ϕ(X) = X ∗ . To this end, take a Cauchy sequence
{pn } representing an arbitrary element of X ∗ . Then since X is complete, {pn } → p for some p ∈ X. The
constant sequence {p} is then equivalent to {pn } as d(pn , p) → 0 as n → ∞. Thus, the class of {pn } in X ∗
is in the image of ϕ as desired.