What does it mean for a function to be continuous? An elementary calculus course would define:

**Definition 1:** Let
and
be a
function. Let
and
. The function
has limit
as x approaches a if for every
,
there is a
such that for every
with
,
one has
. This is expressed as

**Definition 2:**The function f is said to be continuous at
if

On the other hand, in a first topology course, one might define:

**Definition 3:** A topological space is a pair (X,
) where
X is a set and
is a collection of subsets of X (called the open
sets of the topological space) such that

- The Union of any number of open sets is an open set.
- The Intersection of a finite number of open set is an open set and.
- Both X and the empty set are open.

As an abbreviation, we speak of the topological space X when we don't need to refer to . A set is closed if its complement is open.

**Exercise 1:** If (X,
) is a topological space and
,
then (A,
) is also a topological space.
We say that this is the topology induced on A by the topology on X.

**Definition 4:** A function
of topological spaces
is continuous if for every open subset
of
,
is an open subset
of X.

**Definition 5:**An open rectangle
is defined to be a subset of
of the form
, i.e. There are real numbers
and
where
for all i
and

In the case of
, we can define a topology by saying that a subset
of
is open if every point
is contained in an open rectangle
which is entirely contained in
. We call this the *usual* topology of
and for any subset
of
, the induced topology is called the
*usual* topology of
.

**Exercise 2:** Make a reasonable definition of closed rectangle. Prove
that open rectangles are open sets and closed rectangles are closed sets.

**Exercise 3:** Instead of rectangles, we could have used open balls.
Show that the resulting topology would be the same.

We are now ready for:

**Proposition 1:** With the usual topology on
and
, the two
notions of continuous function are equivalent.

**Proof:** This is easy if you did Exercise 3.

The most important theorem in one variable calculus is:

**Theorem 1:** (Mean Value Theorem) If
(where
) is continuous on
and differentiable on
, then
there is at least one
satisfying

The principal ingredient of the proof of this theorem is:

**Theorem 2:** (Extreme Value Theorem) Every continuous function
(where
has at least one absolute
minimum and at least one absolute maximum.

Usually, in an elementary calculus course, one proves the Mean Value
Theorem, but not the Extreme Value Theorem -- the second result is usually
taken to be *intuitively* obvious. Actually, this is a rather subtle
and difficult result. To begin with, let us note that in fact, the
Extreme Value Theorem holds when the interval
is replaced with
a closed and bounded subset of the real numbers, where bounded is defined as:

**Definition 6:** A subset
of
is said to be
bounded if it is contained ins ome open rectangle.

**Exercise 4:** Find counterexamples to Theorem 2 if you either just
assume that the interval is not closed or that it is not bounded.

**Definition 7:** A cover for a subset
of a topological space X
is simply any collection
of subsets of X whose union contains
. The cover is called open (respectively finite) if the subsets are all
open (respectively are finite in number).

**Definition 8:** A subset A of a topological space X is said to be
compact if every open cover of A contains a finite subcover (i.e. a finite
subset of the cover is itself a cover).

**Proposition 2:** If
is continuous and
is compact, then so is
.

**Proof:** If
is an open cover of
, then
is an open cover of A (Why?).
Since
is compact,
has a finite subcover,
say
, where the
are all in
.
Now, one can verify that
form a finite cover of
.

**Exercise 5:** With the usual topology on
, if
is compact, then
is both closed and bounded.

**Theorem 5:** (Heine-Borel Theorem) With the usual topology on
,
a subset
of
is compact if and only if it both closed and bounded.

**Note:** The Extreme Value Theorem follows: If
is continuous, then
is the image of a compact set and so is compact
by Proposition 2. So, it is both closed and bounded by Exercise 5. In
particular, the least upper bound of
(which exists by the
completeness property of the real numbers) is in
(Why?). So,
has an absolute maximum.

The Heine-Borel Theorem can be proved in at least two ways. The first method proceeds by building up larger and larger sets which are known to be compact. One starts with Lemma 1 below and then uses Lemma 2 to inductively conclude that any closed rectangle is compact. Finally, Lemma 3 completes the proof.

**Lemma 1:** A closed interval
is compact.

**Proof:** Let
be an open cover of
(assumed
non-empty). The set
of all
such that
contains
a finite subcover of
is bounded. So it has a least upper bound, say
.
Let
contain
and hence an open interval
containing
.
Choose
in this open interval. Then
and so there is a finite
subcover
of
. But then
is a finite subcover of
for
any
contrary to the choice of
.

**Lemma 2:** If
and
are compact, then so is
.

**Proof:** Suppose
is an open cover of
.
If
,
is compact. Further,
is also an open cover of
and so this set has a finite subcover
.
For every
, there is an open rectangle
containing
and contained
in some one of the
; the set of these is rectangles is an open cover of
and so admits of an open subcover
of the same set. Each rectangle
where
and
are open
rectangles. Letting
, it follows that
is actually a finite open cover of
.

For each , we can do the construction of the last paragraph. The set of all the for all the is an open cover of and so it admits of a finite subcover . But then the set of all the 's associated with all the 's is a finite subcover of .

**Lemma 3:** If every rectangle is compact, then every closed and bounded
subset of
is compact.

**Proof:** If
is closed and bounded, then it is contained in
a closed rectangle
. If
is an open cover of
, then adding
the complement of
in
to the cover gives an open cover of
. Since
is compact, there is a finite subcover. But then after possibly removing
the complement of
in
, we get a finite subcover of
.

The second proof of the Heine-Borel Theorem proceeds by binary search. We will prove it in the case an interval and leave the generalization to a rectangle as an exercise.

**Proof:**(Heine-Borel, case where
) Suppose the result is false. Then there
is an open cover
of
without a finite subcover. Let's look for
smaller intervals with the same property with respect to this fixed cover.
Clearly, either the left or the right half of
must not have a finite
subcover (Otherwise, the union of the subcovers for each half would give a
subcover for the whole interval.) We can now apply the same reasoning to
a half that does not have a finite subcover, etc. This gives a sequence
of intervals

where for every and such that each subinterval does not admit of a finite subcover of . Clearly, one has . Let be the least upper bound of the set of these . Then is in and so there is a set in with . It is now easy to see that contains almost all the intervals , which is a contradiction.

**Exercise 6:** Generalize the above proof to show that every closed
rectangle in
is compact.

**Definition 9:** The derivative of the function
is
whenever it exists.

**Proposition 3:** With
as in the definition, suppose that the
derivative of
at
exists. If
is a local maximum or local minimum
of
, then the derivative of
at
is zero.

**Proof:**Suppose, for example, that
is a local maximum of
,
i.e.
for all
in an open interval containing
. But then
for
in this interval with
, one has
and so the limit as
approaches
from the right must be non-positive.
By taking
in the interval with
, one has
and so the limit as
approaches
from the left must be non-negative. If
the limit exists, then these two must be equal, and so the derivative is zero.
The proof in case of a local minimum is analogous.

**Proposition 4:** (Rolle's Theorem) If
is continuous on
and differentiable on
and
, then
for some
, one has
.

**Proof:** If the result is false, then Proposition 3 says that there
is no local maximum or local minimum in
. On the other hand, the
Extreme Value Theorem says that there is at least one absolute maximum and
at least one absolute minimum; so both must be at the endpoints. But, then
the function must be constant in the entire interval. This is a contradiction.

We can now prove:

**Corollary 1:** (Mean Value Theorem) If
and
are both real valued
functions continuous on
and differentiable on
and if the
graphs of
and
intersect at
and
, then
there is at least one
satisfying
.

**Proof:** Apply Rolle's Theorem to the function
.

**Note:** Theorem 1 is the special case where
is the straight
line from
to
.

**Corollary 2:** (Taylor's Theorem with Remainder) Let
and its first n derivatives be continuous on
and
.
Then for all
, one has:

for some .

**Proof:** Consider the function:

where the constant A is chosen so that . The derivative can be calculated using the product rule:

Using Rolle's Theorem, we know that there is a satisfying ; this is the we are looking for.