Math 597. Homework 1 Bryden Cais

(1) Suppose that X = {α ∈ S : f(α) > 0} is uncountable and define X_{n} = {α ∈ S : f(α) > 1/n}. Then clearly

X = ⋃_{n=1}^{∞} X_{n}.

If each X_{n} were finite, then X would be a countable union of finite sets and hence countable. It follows that there exists n_{0} such that X_{n_{0}} is infinite. Let T_{r} be any finite subset of X_{n_{0}} consisting of r elements (such a subset exists for each r > 0 since X_{n_{0}} is infinite). We then have

∑_{α∈S} f(α) ≥ ∑_{α∈X_{n_{0}}} f(α) ≥ ∑_{α∈T_{r}} \frac{1}{n} = \frac{r}{n}

for every positive integer r. It follows that ∑_{α∈S} f(α) ≮ ∞, which is the contrapositive.

(2) Let x, y ∈ E. Without loss of generality, we may assume that x < y. Suppose that [x, y] ⊈ E. Then there exists some z ∈ R \ E with x < z < y. Observe that since

a(y) = sup{u ∈ R \ E : u < y}

we have a(y) ≥ z. Similarly, since

b(x) = inf{u ∈ R \ E : x < u}

we have b(x) ≤ z. Thus, b(x) ≤ a(y) Trivially, we have a(x) < b(x) and a(y) < b(y) so that (a(x), b(x)) ∩ (a(y), b(y)) = ∅. Therefore, if (a(x), b(x)) ∩ (a(y), b(y)) ≠ ∅ then [x, y] ⊆ E. It follows that

{u ∈ R \ E : u < y} = {u ∈ R \ E : u < x}

and similarly that

{u ∈ R \ E : x < u} = {u ∈ R \ E : y < u},

for otherwise, in any case, we would obtain some z ∈ R \ E with x < z < y, a contradiction. This clearly implies that a(x) = a(y) and b(x) = b(y).

(3) Let f : E → C and g : E → [0, 1] be given by

f(\{ϵ_{n}\}) = ∑_{n=0}^{∞} \frac{2ϵ_{n}}{3^{n}}

g(\{ϵ_{n}\}) = ∑_{n=0}^{∞} \frac{ϵ_{n}}{2^{n}} .

For {ϵ_{n}}, {ϵ̃_{n}} ∈ E, set m = min{n : ϵ_{n} ≠ ϵ̃_{n}}.

(a) We have

f(\{ϵ_{n}\}) − f(\{ϵ̃_{n}\}) = ∑_{n=0}^{∞} \frac{2(ϵ_{n} − ϵ̃_{n})}{3^{n}} = ∑_{n=m}^{∞} \frac{2(ϵ_{n} − ϵ̃_{n})}{3^{n}}.

Since ϵ_{n}, ϵ̃_{n} ∈ {0, 1} we have |ϵ_{n} − ϵ̃_{n}| ≤ 1 so that

| ∑_{n=m}^{∞} \frac{2(ϵ_{n} − ϵ̃_{n})}{3^{n}} | ≤ ∑_{n=m}^{∞} \frac{2}{3^{n}} = \frac{1}{3^{m−1}} .

On the other hand, the triangle inequality gives

| ∑_{n=m}^{∞} \frac{2(ϵ_{n} − ϵ̃_{n})}{3^{n}} | ≥ \frac{2}{3^{m}} − | ∑_{n=m+1}^{∞} \frac{2(ϵ_{n} − ϵ̃_{n})}{3^{n}} | ,

and as we have just seen, the second term above is at most 1/3^{m}. It follows that we have the bounds

\frac{1}{3^{m}} ≤ |f(\{ϵ_{n}\}) − f(\{ϵ̃_{n}\})| ≤ \frac{1}{3^{m−1}}

for any two different sequences ϵ_{n}, ϵ̃_{n}. Moreover, these bounds are sharp as the sequences

ϵ_{n} = \begin{cases}1 & if n = m \\ 0 & otherwise \end{cases}  ϵ̃_{n} = \begin{cases}0 & if n ≤ m \\ −1 & otherwise \end{cases}

and

ϵ_{n} = \begin{cases}1 & if n ≥ m \\ 0 & otherwise \end{cases}  ϵ̃_{n} ≡ 0

attain the lower and upper bounds, respectively.

(b) Observe that we obviously have |f({ϵ_{n}}) − f({ϵ̃_{n}})| ≥ 0 and that

∑_{n=m+1}^{∞} \frac{1}{2^{n}} = \frac{1}{2^{m}} ,

so that this bound is sharp. By the same estimates in part (a), we find

| ∑_{n=m}^{∞} \frac{(ϵ_{n} − ϵ̃_{n})}{2^{n}} | ≤ ∑_{n=m}^{∞} \frac{1}{2^{n}} = \frac{1}{2^{m−1}} ,

so that we have the bounds

0 ≤ |g(\{ϵ_{n}\}) − g(\{ϵ̃_{n}\})| ≤ \frac{1}{2^{m−1}} .

Moreover, the upper bound is sharp also, as it is attained by the sequences

ϵ_{n} = \begin{cases}1 & if n ≥ m \\ 0 & otherwise \end{cases}  ϵ̃_{n} ≡ 0.

(c) Replacing x, y by f(x), f(y) (recall that f is bijective) we find that such a constant C exists if and only if there is a constant C with

|\frac{g(\{ϵ_{n}\}) − g(\{ϵ̃_{n}\})}{f(\{ϵ_{n}\}) − (\{ϵ̃_{n}\})}| ≤ C

for all {ϵ_{n}}, {ϵ̃_{n}} ∈ E. But letting

ϵ_{n} = \begin{cases}1 & if n ≥ m \\ 0 & otherwise \end{cases}  ϵ̃_{n} ≡ 0,

we have

|\frac{g(\{ϵ_{n}\}) − g(\{ϵ̃_{n}\})}{f(\{ϵ_{n}\}) − (\{ϵ̃_{n}\})}| = \frac{3^{m−1}}{2^{m−1}} ,

which is unbounded as m → ∞ so that no such constant C independent of m can exist.

(d) Yes. By parts (a) and (b) we have

\frac{1}{3^{mη}} ≤ |f(\{ϵ_{n}\}) − f(\{ϵ̃_{n}\})|^{η}

for any η > 0 and

|g(\{ϵ_{n}\}) − g(\{ϵ̃_{n}\})| ≤ \frac{1}{2^{m−1}} ,

so that we have

\frac{|g(\{ϵ_{n}\}) − g(\{ϵ̃_{n}\})|}{|f(\{ϵ_{n}\}) − (\{ϵ̃_{n}\})|^{η}} ≤ 3^{η} (\frac{3^{η}}{2} )^{m−1} .

thus, for any 0 < η < log(2)/ log(3) we have 3^{η}/2 < 1 so that the right side above is bounded for all m > 0, as required.