If you want to drive along Interstate 95 from the George Washington Bridge to the Connecticut state line, you must pass through a sliver of Manhattan, then a trek through The Bronx, then Pelham Manor, then New Rochelle, then Larchmont, then Harrison, then Rye, and finally through Port Chester. When you are almost through New Rochelle, you can't simply decide, "I never liked Larchmont, so let's just skip it," and jump through to Harrison (which would, coincidentally, save you a buck at the toll plaza). If you stay on I95 from the bridge to Connecticut, you must visit all the places in between.

Likewise, if your 20,000 gallon swimming pool starts out empty, and you begin filling it, there will be a moment when it contains exactly 5,673 gallons. There will be another moment when it contains exactly 12,891 gallons. No matter what number of gallons you name between zero and 20,000, there will be a moment when your pool contains exactly that many gallons. It doesn't matter how hard you turn on the hose, or how many times you adjust the valve. You just don't get to skip any numbers, no matter how unlucky your numerologist says they are.

Just one more example -- the rubber-burning monster of a sports car you just bought can go from 0 to 60 mph in 6.9 seconds. But it can't get you from 0 to 60 without at some time taking you through all the speeds in between.

All of the above are examples of nature's tendency to be *continuous*.
Nature doesn't like instantaneous changes or perfectly sharp edges.
Nature wants things to happen smoothly. Even something as abrupt as the
shock wave of an explosion does not happen instantaneously, but builds up
over a period of time, though that time be measured in microseconds.

We can talk about real valued functions of a real variable in the same
way. We have a sense about functions like
`f(x) = x ^{2}`

These notions are useful for getting an idea of what continuity is all about, but they lack the rigor we need to do mathematics with them. There simply is no mathematical definition of smooth or sharp or of a pencil. Fortunately, the concept of limits, as we have developed it so far, gives us the perfect tool to nail down the notion of continuity.

First notice that continuity is a local property. If Interstate 95 has a bridge that is out, then we would like to say that it is continuous everwhere except at the missing bridge. The same is true when we talk about functions.

Suppose we define the function, ` u(x) = 0 `` x < 0 `` u(x) = 1 `` x > 0 ``u(x)` is called the unit step function,
and is graphed in figure 3-1).
You can define it to be whatever you like
at ` x = 0``u(x)` is smooth and
continuous everwhere except at ` x = 0`*at*
` x = 0``u(x)` takes an
instantaneous jump -- that is that it goes from 0 to 1 without passing
through all the points in between. So we have a notion that `u(x)`
is *discontinuous* at ` x = 0`

So what is it that's happening to `u(x)` everwhere except at
` x = 0 ``a`, on the `x` axis.
Except at ` a = 0`

lim u(x) = u(a) eq. 3.0-1 xAt every~~> a~~

The function, `f(x) = x ^{2}`

But first, let me answer a few questions that curiosity may be stuffing
into your head
at this very moment. In the example of `u(x)` we saw a function that
was continuous everywhere except at one point. Is there a function that is
continuous everywhere except at several points? Yes. Is there one that
is continuous
everywhere except at infinitely many isolated points? Yes. Is there one
that is
continuous nowhere at all? Yes. Is there one that is continuous only at
a single point? Yes. Is there a function that is continuous everywhere
yet cannot be graphed? Yes. And later on I shall give you examples of
all of these.
**optional material** right now. Or you can continue reading and you will
get to it eventually.

But now let's get back to our analysis of
` f(x) = x ^{2}`

lim xand we say that this holds for any^{2}= a^{2}eq. 3.0-2 x~~> a~~

|xwhenever^{2}- a^{2}| £ e eq. 3.0-3

But can we prove that 3.0-2 holds for all `a` using the contract?
Let `error = x - a``error` is how far `x` is from `a`, and we will express
our contract using that instead of using `x`. So
` |error| £ d `` x = a + error``x` into 3.0-3, and the
contract looks like:

|(a + error)whenever^{2}- a^{2}| £ e eq. 3.0-4

|2a error + errorwhenever^{2}| £ e eq. 3.0-5

Can we choose `error`
so that 3.0-5 always holds? Certainly. Just choose `error` so that
it is the lesser of ` |e /(4a)| `` sqrt( e /2 )``error` gets closer to zero.
You want to pick the candidate that gives you the smallest possible sum.
If you do that, you guarantee that
both halves of the sum in 3.0-5 are are positive and less than or equal
to `e /2`, and hence the sum is less than or equal to
`e`.
And clearly, if you choose `error` to be even
closer to zero, you are still guaranteed that 3.0-5 holds. But the point
is, if you choose `error` no worse than we have chosen it, success
is guaranteed.

That works for all `a`, except zero.
But when
` a = 0``a` is zero, simply choose `error`
to be ` sqrt( e /2 ) `

So now you have a recipe for finding `error` given `e`
and `a`. Just make `d` the same as the `error` we
chose in the previous two paragraphs, and
you have a recipe for finding `d` given `e` and
`a`. And that is what the contract said we should be able to do.
Not only that, but it works for any `a`.

You are likely to be called upon on an exam to prove continuity of some
function using the method just detailed, or one similar to it. Read over
the discussion of why
` f(x) = x ^{2} `

So what about the first function we looked at, our friend, `u(x)`?
If it really is continuous everywhere except at zero, then our limit test
should pass everwhere *except* at zero, and there it should fail.

Our contract for `u(x)` is: for any
` e > 0 ``d` such that

|u(x) - u(a)| £ e eq. 3.0-6whenever

Whenever `a` is not zero, we can always find a `d` small
enough so that the interval that extends out a distance `d` on
each side of `a` does not include zero. You can see this graphically
on a number line.

0 <No matter how close~~d~~~~>a<~~~~d~~~~> --------------------------|-------[---------------+----------------]------- fig. 3-2~~

Well, whenever `x` is inside
the interval shown, it is certainly true that
` |x - a| £ d``x`
is in the interval is exactly the same as saying
` |x - a| £ d`

|x - a| £ dAnd since the entire interval is on the same side of zero,Ûx is in the interval eq. 3.0-7

That means that whenever ` |x - a| £ d`` u(x) = u(a)`` |u(x) - u(a)| = 0``e`. So when `a` is not zero, our
recipe for finding a `d` is simply, choose `d`
small enough so that the interval it creates around `a` does not
include zero. If you do that, the contract holds.

But what about when ` a = 0``u(x)` is
truly not continuous at this point, then the contract should always fail
here. First, recollect that I said you could choose `u(0)` any
way you like, including leaving it undefined. Well, if you leave it
undefined, then `u(x)` can't possibly be continuous at zero because
our limit definition of continuity requires a function to be defined
wherever it is
continuous. Refer back to 3.0-1 to see this. How can we compare `u(x)`
with `u(0)` if `u(0)` doesn't even exist?

Some candidates we might pick for `u(0)` are 0, 1, or anything in
between. It turns out, it doesn't matter. You can pick any real number
you like and say that that is what `u(0)` equals, and the contract
will still fail at
` a = 0`

At ` a = 0`` |x| £ d`` d > 0 ``x` into ` |x| £ d`

<Again,~~d~~~~>0<~~~~d~~~~> -----------------------[-----------|-----------]--------------------- fig. 3-3~~

|x| £ dThere simply is no way to exclude zero from this interval. More important than that, the intervalÛx is in the interval eq. 3.0-8

So if you told me that you wanted `u(0)` to be 0, then there are
values of `x` in the interval shown for which
` u(x) = 1 `` 0 < x £ d`

|u(x) - u(0)| = 1 eq. 3.0-9That means that if you choose any

Similarly, if you say that
` u(0) = 1``x` in the
interval for which ` u(x) = 0 `` -d ³ x > 0`` e < 1``d` that makes the contract hold.

And suppose you pick `u(0)` to be something that isn't 0 or 1? Well,
let's do that. Let's call it `y`.
We still have a problem.
`y` is different from any `u(x)` when `x` is on the
positive side of the interval (where
` u(x) = 1``y` is different from any
`u(x)` when `x` is on the negative side of the interval
(where ` u(x) = 0`` |1 - y| `` |0 - y| ``e` that is closer
to zero than that, and you guarantee that I will never be able to find
a `d` close enough to zero to make the contract hold. No matter
how small I make `d`, the interval will include an `x`
for which `u(x)` is farther from `y` than `e` is
from zero.

I just can't win, no matter what I choose for `u(0)`. The contract
always fails. And that's why `u(x)` cannot be continuous at
`x = 0`

We began this discussion with the metaphor of a highway, and that to
get from point A to point B on a highway, you have to visit all the
points in between. We went on to use a definition involving limits
to define what continuity of a function is. Specifically, we stated
that a function `f(x)` is *continuous* at
`x = a``f(a)` exists and

f(a) = lim f(x) eq. 3.0-10 xBut we never made the connection from the definition to the points-in-between concept. The~~> a~~

The usual statement of the intermediate value theorem is: if `f(x)`
is continuous on the closed interval,
` a £ x £ b ``f(x)` is continous
at every point on the interval and the interval includes the
endpoints), and `f(a)`
is not equal to `f(b)`, then for every value, `y`, that
falls in between `f(a)` and `f(b)`, there exists at least
one point,
`c`, in between `a` and `b` such that
` f(c) = y`

Read that statement of the intermediate value theorem over several times and make sure you understand what it means. You are likely to be tested on it. Try to see why the theorem implies, among other things, that if

f(x) = xis continuous everywhere and^{2}

Think of it this way. If you take a pencil to paper and begin drawing a mark always moving from left to right, but while you do that you also move the pencil-point up and down, and you never pick the pencil-point off the paper, what can you say about the mark you made? If you made it to the top of the page anywhere, and you made it to the bottom of the page anywhere, then you made it to all the heights in between. It's just common sense.

In fact, the intermediate value theorem seems intuitively obvious until
somebody comes along and says, "If it's so obvious, then prove it."
It is then that you realize that its everyday applications to our
experiences in life are obvious, but a rigorous proof is another matter.
The proof is not easy, and the likelihood that you will be asked to
produce it in a first year calculus course is almost nil. So it is
given here as **optional material** for the curious and the brave.
This proof is not so hard that
a first year student can't understand it. So if you feel up to a challenge,

The intermediate value theorem has a partner that also seems obvious
called *The extreme value theorem*.
It is stated without
proof in almost every first year calculus text. This theorem says if
`f(x)` is continous on the closed interval,
` a £ x £ b ``f(x)` is continous
at every point on the interval and the interval includes the
endpoints), then `f(x)` is *bounded* for all `x`
on the interval. So there is some value that `f(x)` never
exceeds and another that `f(x)` never gets under.
But the extreme value theorem is stronger than just that.
It also says that over all the `x`'s on the closed
interval there is a value of `x` that gives you a maximum `f(x)`
and another that gives you a minimum `f(x)`.

All this is saying is that if you fasten one end of a clothes line to one pole and the other to another pole, it doesn't make any difference how much slack you leave or how many branches you drape it through, there will be a highest point on the clothes line and a lowest point on the clothes line. This is true because 1) the clothes line is continuous and 2) because the clothes line includes its endpoints.

In other words, a function that is continuous on a closed interval
*cannot* run off to infinity or to minus infinity on that
interval. It must remain always between some minimum value and
some maximum value. Not only that, there must be a point, `c`,
on the interval such that `f(c)` is the maximum value, and
another point, `d`, on the interval such that `f(d)`
is the minimum.

It is important that the interval be closed -- that is that it include the endpoints. If the interval does not include the endpoints, it is easy to come up with a counterexample. The function

x f(x) =is continuous on the open interval (that is excluding the endpoints) that runs~~eq. 3.0-11 x~~^{2}- 1

The proof of this theorem is even harder than that of the intermediate
value theorem, and you won't be asked to produce it in a first year
course. But it will be assumed that you know what the theorem means.
It plays a big role in the theory of maximums and minimums. For the
really brave, and because I hate to give you something as important
as this without proof, you can see the proof as **optional material**
by

The problems that follow are proofs. You should know that proofs concerning continuity may be on the exam.

Now here is one that you should be able to prove, and that you might
encounter on an exam. Suppose `f(x)` is continuous on the
closed interval ` a £ x £ b``c`, on that interval (and not an endpoint)
for which
` f(c) > 0``d`
such that ` f(x) > 0 ``x` in the
interval,
` c - d < x < c + d``delta-epsilon`
contract that implies. **Hint:** Make `e` *less than*
`f(c)` but still positive.

As always, make a sincere effort to work this problem on your own. If
you give up or if you want to compare your proof with mine,

Here's another theorem (which I call *The Stepping Stone Theorem*)
that you ought to be able to work through at least part of.
Prove that a function, `f(x)`, is continuous at a point,
` x = c`` x _{1}, x_{2}, x_{3}, ... `

Notice that this is an "if or only if" proof. That means it is really
two theorems in one. One theorem says that if for *every* set
of stepping stones that leads to the river bank, the moss on the
stepping stones leads to the moss on the riverbank, *then* the
moss function is continuous at the riverbank. The other theorem is
that if the moss is continuous at the riverbank, *then* for every
set of stepping stones leading to the riverbank, the moss on them
leads to the moss on the riverbank.

The usual approach to an "if and only if" theorem is to divide it into its two constituent theorems and prove each one separately. And that is what I recommend you do here. The second part of this one (that is the "only if" part) is the easier one to prove, so perhaps you should start with that. This part asserts that if the moss function is continuous at the riverbank, then each set of stepping stones that leads to the riverbank has moss that leads to the moss on the riverbank.

Approach it by demonstrating
that `f(x _{n})` can always be brought to within

Go ahead and prove the "only if" part, and I'll give you a pass
on doing the more difficult "if" part. When you are done, you
can look at both proofs by

email me at *hahn@netsrq.com*