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Lecture 18: More continuity
Let us begin with some examples.
1. Not every continuous function on a non compact set is uniformly continuous. If E
is any non-closed subset of R then there exists a continuous function on E that is
both unbounded and not uniformly continuous. Take x0 to be any limit point of E
that is not in E. Then f (x) = (x − x0 )−1 is continuous but unbounded. Further f
is not uniformly continuous because there is no δ > 0 such that for all x, y ∈ E with
|x − y| < δ we have |f (x) − f (y)| < 1. If there were, we could just choose some y ∈ E
with |y − x0 | < δ/2 and then deduce that all points z within distance δ of y have
f (z) ≤ f (y) + 1. But this is impossible.
2. If E is an unbounded subset of R then there is an unbounded continuous function on
E: just take f (x) = x.
3. The only polynomials that are uniformly continuous on all of R are those of degree at
most 1. Indeed, take f (x) = an xn + · · · a1 x + a0 with an 6= 0 and n ≥ 2 and assume
that there exists δ > 0 such that if |x − y| < δ then |f (x) − f (y)| < 1. Then consider
points x, y of the form x, x + δ/2: you can check that
{|f (x) − f (x + δ/2)| : x ∈ R}
is unbounded, giving a contradiction. (The problem here is that for fixed δ, the quantity
|f (x) − f (x + δ/2)| grows to infinity as x → ∞. This is not the case if n = 0 or 1.)
There are other ways that functions can fail to be uniformly continuous. We will see later,
however, that any differentiable function with bounded derivative is uniformly continuous.
Connectedness and the IVT
We would like to prove the intermediate value theorem from calculus and the simplest
way to do this is to see that it is a consequence of a certain property of intervals in R.
Specifically, an interval is connected. The definition of connectedness is somewhat strange so
we will try to motivate it. Instead of trying to envision what connectedness is, we will try to
capture what it is not. That is, we want to call a metric space disconnected if we can write
it as a union of two sets that do not intersect. There is a problem with this attempt at a
definition, as we can see by considering R. Certainly we can write it as (−∞, 1/2) ∪ [1/2, ∞)
and these sets do not intersect, but we still want to say that R is connected. The issue in
this example is that the sets are not separated enough from each other. That is, one set
contains limit points of the other. This problem is actually resolved if we require that both
sets are open. (But you have to think about how this resolves the issue.)
Definition 0.1. A metric space X is disconnected if there exist non-empty open sets O1 and
O2 in X such that X = O1 ∪ O2 but O1 ∩ O2 = ∅. If X is not disconnected we say it is
Connectedness and continuity also go well with each other.
Theorem 0.2. Let X, Y be metric spaces and f : X → Y be continuous. If X is connected
then the image set f (X), viewed as a metric space itself, is connected.
Proof. As stated above, we view f (X) ⊂ Y as a metric space itself, using the metric it
inherits from Y . To show that f (X) is a connected space we will assume it is disconnected
and obtain a contradiction. So assume that we can write f (X) = O1 ∪ O2 with O1 and O2
nonempty, disjoint, and open (in the space f (X)). We will produce from this a disconnection
of X and obtain a contradiction.
Now consider U1 = f −1 (O1 ) and U2 = f −1 (O2 ). These are open sets in X since f is
continuous. Further they do not intersect: if x is in their intersection, then f (x) ∈ O1 ∩ O2 ,
which is empty. Last, they are nonempty because, for example, if y ∈ O1 (which is nonempty
by assumption) then because O1 ⊂ f (X), there exists x ∈ X such that f (x) = y. This x is
in f −1 (O1 ).
So we find that X is disconnected, a contradiction. This means f (X) must have been
• Let X be a discrete metric space. If X consists of at least two points then X is
disconnected. This is because we can let O1 = {x} for some x ∈ X and O2 = O1c . All
subsets of X are open, so these are open, disjoint, nonempty sets whose union is X.
• Every interval in R is connected. You will prove this on the homework.
Theorem 0.3 (Intermediate value theorem). Let f : [a, b] → R for a < b be continuous.
Suppose that for some L ∈ R,
f (a) < L < f (b) .
Then there exists c ∈ (a, b) such that f (c) = L.
Proof. Since f is continuous, the space f ([a, b]) is connected. Since
O1 := (−∞, L) ∩ f ([a, b]) and O2 := (L, ∞) ∩ f ([a, b])
are both nonempty (because f (a) ∈ O1 and f (b) ∈ O2 ), open in f ([a, b]), and disjoint, it
cannot be that their union is equal to f ([a, b]). Therefore L ∈ f ([a, b]) and there exists
x ∈ X with f (x) = L.
Let us spend a couple of minutes on types of discontinuities for real functions. Let E ⊂ R,
f : E → R and x0 ∈ E. (Draw some pictures.)
• x0 is a removable discontinuity of f if limx→x0 f (x) exists but it not equal to f (x0 ).
• x0 is a simple discontinuity of f if limx→x−0 f (x) exists, as does limx→x+0 f (x), but they
are not equal. Here the first limit is a left limit; that is, we are considering f as being
defined on the metric space E ∩(−∞, x0 ] and taking the limit in this space. The second
is a right limit, and we consider the space as E ∩ [x0 , ∞). This corresponds to saying,
for example, that
lim− f (x) = L
if for each > 0 there exists δ > 0 such that if x0 − δ < x < x0 then |f (x) − L| < .
• x0 can be a discontinuity but not captured above. Consider f : R → R given by
sin(1/x) if x 6= 0
f (x) =
if x = 0
Here there is not even a limit as x → 0. This is because we can find a sequences (xn )
converging to 0 such that (f (xn )) does not have a limit. Take
xn = 2/(nπ) .
Derivatives: Ch. 5
Continuous functions are nicer than most functions. However we have seen that they can still
be rather weird (recall the function that equals 1/q at a rational expressed in lowest terms
as p/q). So we move on to study functions that are even nicer, and for this we henceforth
restrict to functions from R to R. We could start at the very bottom, first studying constant
functions f (x) = c and then linear functions f (x) = ax + b, then quadratics, etc. But I trust
you learned about these functions earlier. Noting that constant functions are just special
cases of linear ones, we set out to study functions that are somehow close to linear functions.
The idea we will pursue is that even if a function f is wild, it may be that very close to
a particular point x0 , it may be well represented by a linear function. For a good choice of
a linear function L, it would make sense to hope that
lim (f (x) − L(x)) = 0 .
If f is already continuous then this is not much of a requirement: we just need L(x0 ) = f (x0 ).
So this just means that L(x) can be written as L(x) = a(x − x0 ) + f (x0 ).
We will look for a stronger requirement on the speed at which this difference converges
to zero. It should go to zero at least as fast as x − x0 does (as x → x0 ). In other words, we
will require that
f (x) − L(x)
= 0 or in shorthand, f (x) − L(x) = o(x − x0 ) .
x − x0
Plugging in our form of L, this means
f (x) − f (x0 )
−a =0 ,
x − x0
f (x) − f (x0 )
x − x0
Rewriting this with the notation above, we get
f (x) = f (x0 ) + a(x − x0 ) + o(x − x0 ) ,
or setting x = x0 + h,
f (x0 + h) = f (x0 ) + ah + o(h)
as h → 0. Again, the symbol o(h) represents some term such that if we divide it by h and
take h → 0, it goes to 0.
Definition 0.4. Let f : (a, b) → R. We say that f is differentiable at x0 ∈ (a, b) if
f (x) − f (x0 )
exists .
x − x0
In this case we write f 0 (x0 ) for the limit.
Proposition 0.5. Let f : (a, b) → R be differentiable at x0 . Then f is continuous at x0 .
lim f (x) = f (x0 ) + lim [f (x) − f (x0 )] = f (x0 ) + lim
f (x) − f (x0 )
lim (x − x0 ) = f (x0 ) .
x − x0
The converse is not true. Consider the function f : R → R given by f (x) = |x|. Then
f (x) = max{x, 0} − min{x, 0} ,
so it is continuous. However, trying to compute the derivative at x = 0, we get
x→0 x
which does not exist (it has a right limit of 1 and left limit of -1).
We will now play the same game as we did for continuity, trying to find which functions
are differentiable. Here are some examples.
1. f (x) = x:
f (x) − f (x0 )
x − x0
So f 0 (x) exists for all x and equals 1.
2. f (x) = xn for n ∈ N:
xn − xn0
(x − x0 )(xn−1 + xn−2 x0 + · · · + xxn−2
+ x0n−1 )
= lim
x − x0
x − x0
= lim x
+ · · · + x0
= nx0n−1 .
So f 0 (x) exists for all x and equals nxn−1 .