1. Let be a sequence of open sets containing Q. Prove that contains an irrational number.

Label the rationals q_{1}, q_{2}, …, and construct inductively a nested sequence

of closed intervals with the property that

and Start by taking I_{1} an interval containing q_{1} and contained in

U_{1} (this is made possible by the assumption that U_{1} is an open neighbor-

hood of q_{1}), having constructed I_{n}, take an open interval J contained in

I_{n} and not containing q_{n}, and take for I_{n+1} any closed interval contained

in the open set which is nonempty since

is an open neighborhood of Q, and

It is manifest thatand
thatfor a descending chain

of (closed) intervals follows from the completeness axioms of R, alternatively,

from the finite intersection property characterization of compact-

ness. Finally,contains no rational number,
since by construction

The conclusion follows.

**Remark.** Some of you noted (a special case of the)
**Baire category**

**theorem:** if X is a complete metric space and U_{n} a sequence of
dense open

subsets, thenis dense. The way this applies
is by taking X := R

and the sequence of dense open subsets: the
intersection of this

sequence must be dense and in particular nonempty, but it contains no

rational number, hence the conclusion. The proof of the general result is

exactly the same.

2. Let f : [0, 1] -> R be a function, and
its graph. Prove or

disprove each of the four implications:

f is continuous is compact is connected:

The only implication that does not hold is that
connectedness of the graph

does not imply compactness, as demonstrated by the example of the topologist's

sine curve (whose connectedness was shown on the second homework):

The other three implications hold. Note, first of all,
that is, by

definition, the image of the compact connected interval [0, 1] under the

map this map is continuous if and only if f
is.

Since the continuous image of [0, 1] is compact and connected, it follows

that continuity of f implies compactness and connectedness of

It remains to show that compactness of
implies continuity of f. For

any function the projection map
onto the

x-axis is a continuous bijection. Recall a very important fact shown in

class: a continuous bijection g : X -> Y from a compact space X is a

homeomorphism, i.e. : Y -> X is also
continuous (Proof: continuity of

means that g is a closed map. To show this,
note that compactness of

X implies that every closed subset is
compact, since the continuous

image of a compact space is compact, it follows that g maps any closed

subset of X onto a compact, hence closed, subset of Y , hence g is a closed

map). In our situation, we conclude that compactness of
implies that

the projection map onto the x-axis is a
homeomorphism, and

this is equivalent to the continuity of f.

3. Prove that if converges, then so does

This is an application of the Cauchy-Schwartz inequality:

Let (whenever you see
and a recurrence with

rational coefficients, always think about
and note that the

recurrence may be rewritten as
This

showsSince the right-hand side converges to

0 and for all n, it follows that
Moreover, as it then

follows that is bounded both below and
above, the last formula implies

5. Let and suppose
satisfies

P(x) = 0. Show that there exists a constant C > 0 such that every rational

number satisfies
Use this to show that the

number is transcendental.

The idea is to estimate |P(p/q)| from above by the mean
value theorem,

and from below by the basic arithmetic estimate
which

is simply a reflection of the fact that P(p/q) is a nonzero rational number

of denominator dividing . For the upper
bound, note that the mean

value theorem provides some between x and
p/q such that

Comparing the two bounds,

we may take

This result is called the** Liouville inequality**, and
yields a criterion of

transcendence: if is such that for every d
there exists a rational

number then
is transcendental, such

numbers are called Liouville numbers. An example is the number

and its approximation by the partial sums

which are rational numbers of denominators
since

it follows that
is

a Liouville, hence transcendental, number.

**Remark.** The Liouville inequality is only sharp for
d≤2. For d > 2, this

basic result is superseded by a deep theorem of Klaus Roth, which states

that the exponent d can be replaced by any number > 2. More precisely,

for every algebraic, the set

is finite.

6. Prove that the function
is finite for all t ∈ R.

Is it bounded?

There is an exact formula for the sum:

This follows upon successively setting in
the

formula and forming the resulting telescoping

sum of identites for n = 1, 2, … ,N. The finiteness of S(t) follows

from this expression: it is bounded above in absolute value by

when and it is identically 0 for
It also follows from this

closed form expression that S(t) is not bounded in t, as seen upon taking

**Remark.** In fact, it is not difficult to find the
exact value of S(t) from

the given closed form evaluation:

7. Prove that if if

As noted by most of you, the argument appears in Rudin,
Theorem 8.2.

be the partial sums, and denote
. We have

showing

for r < 1 that the series converges to the
series

Letting arbitrary, there is an M > 0 such
that n > M yields

the right hand-side is and the inequality
become

Letting
proves the required limit

8. Suppose is
continuous. Prove that

converges to f'(x) uniformly on compact
intervals [a, b]. Give

an example where the convergence is not uniform on R.

On a compact metric space, a continuous function is
uniformly continuous,

thus f' converges uniformly on compact intervals, and hence, for a given

, there exists a
such that

imply For
the mean value theorem implies

the existence of for

this
satisfies , and the uniform continuity
implies

that is,
for all and all

This shows the uniform convergence of
on the compact

intervals [a, b].

The convergence need not be uniform on R, the simplest
examples are