Box 3.0b: Proof of the Intermediate Value Theorem
© 1997 by Karl Hahn
Note that the material on this page is optional.
First year students can almost always count on NOT being tested
on this proof
Theorem:
Again the statement of the intermediate value theorem is: if f(x)
is continuous on the closed interval,
a £ x £ b
(that is 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.
Outline of Proof
The outline of the proof is that we assume that for some y
that lies between f(a) and f(b), there is no
value, c, lying between a and b that
satisifies f(c) = y. So y is a
"missing point." From that we shall show
that there must be a point, x, lying between a
and b where f(x) is discontinuous. That is the
contradiction that proves the theorem, since we began by saying that
f(x) is continuous for all x lying
between a and b.
Remember that y is the missing point, that is it is
one for which f(c) = y has no solution on the
interval. The
strategy will be to pin the missing point down to smaller and smaller
intervals. Then we show that the endpoints of those smaller and
smaller intervals form
Cauchy sequences, and
Cauchy sequences always have limits. The definition of continuity
will then demonstrate that the limit of the Cauchy sequences must
be the solution, c, of f(c) = y. So
the only way y can be missing is for f(x) not
to be continuous.
Proof
Notice that on the closed interval,
a £ x £ b,
there must be some points, x, for which
f(x) > y and others for which
f(x) < y . That follows because there can
be none for which f(x) = y. We know that there
is at least one of each type of point because we know that
y falls in between f(a) and f(b), implying
that a is of the opposite type from b.
We take the closed interval
a £ x £ b and
divide it precisely in half -- with both fragments being closed.
This gives us one closed interval from a to
(a + b)/2, and another from
(a + b)/2 to b. One of the following
three situations must be true:
- All of the points, x, for which
f(x) < y fall in one of the fragments and all of
the points, x, for which
f(x) > y fall in the other fragment.
- The lower fragment contains points of both types.
- The lower fragment contains points of just one type, but the
upper fragment contains points of both types.
We now choose a new closed interval contained in the closed a
to b interval by the following scheme: If the first situation
is the case, then we take the middle "half" of the a to b
segment -- that is we take the piece that runs from one quarter way
through to three quarters way through the a to b
interval. If the second situation is the case, then the new interval
is the lower fragment as described above. If the third situation
is the case, then the new interval is the upper fragment as described
above.
Whichever of the three situations was the case, the new interval must
have the following two properties:
- It is half the length of the original.
- It contains points of both types -- that is x's
for which f(x) < y and x's for which
f(x) > y .
Think about it carefully until you
understand why.
Since the new interval contains both types of points, we can perform the
same operation on it and get yet another interval that is half the length
of the new one and completely contained in it. And it too will
contain points of both types.
Clearly we can keep applying this operation over and over again, getting
a new interval each time, always half the length of the last, always
completely contained
in the last, and always containing both types of points. There is no limit
to how many times you can do this.
This gives us a sequence of closed intervals,
I0, I1, I2, ... ,
where I0 is the original a to b interval.
Each In contains all the intervals that
follow, just like Russian nesting dolls.
Let
a0, a1, a2, ...
be the lower endpoints of the sequence of intervals and let
b0, b1, b2, ...
be the upper endpoints of the sequence of intervals.
Notice that both sequences of endpoints are, as the subscripts grow,
confined to a smaller and smaller range. In fact the range each can
be in is halved with each increase of one in the subscript. And by
halving the range each time, you can make the range as close to zero
as you'd like by taking a sufficiently high subscript. That means that
each sequence of endpoints forms a
Cauchy sequence. And every Cauchy
sequence of real numbers converges to a limit that is also a real number.
Not only that, because the difference between bn and
an can be made as close to zero as you like by
choosing n large enough, both Cauchy sequences must
converge to the same real number. Call that real number c.
Observe that c is contained in every interval,
In, in the list described above. This means
that every interval (and not just the ones in the list) that contains
c as an interior point (an interior point of an interval is
one that is not an endpoint), also contains some of the
In's from the list. We only require that it
contain one of them. So why do we know that every interval containing
c as an interior point contains an In?
Well, you can find an In as small as you like
simply by choosing n large enough. That means that you can
always find one that is small enough to fit completly into any interval
containing c.
The consequence of all that is that every interval containing
c as an
interior point, no matter
how small, must
contain both types of points -- those for which
f(x) < y and those for which
f(x) > y .
Why? Because every
every interval containing c as
an interior point also contains an In,
and every In contains points of both types.
We also know, because f(x)
is continuous, that f(c) exists.
In the final stage of this proof, we show that it must be true that
f(c) = y, where y is the missing point
we talked about at the beginning. This is because f(x) is
continuous, and that means,
f(c) = lim f(x)
x > c
And that implies a delta-epsilon contract.
For if f(c) differed from
y then you could set
0 < e < |f(c) - y|.
Because every interval,
no matter how small, that contains c has both types of points
in it (that is the f(x)'s fall on both sides of y),
you could never find a d such that
|f(x) - f(c)| £ e whenever
|x - c| < d.
Hence |f(x) - f(c)| could never be guaranteed
to be less than |f(c) - y|. You would
always be fighting against some nearby point, x, that
causes f(x) to fall on the opposite side of y
from f(c). Why? Because
|x - c| < d
represents an interval that contains c, and
every interval containing
c has x's in it that make f(x) fall on
both sides of y. If one f(x) is on the same
side of y as f(c), then another x very close
by (indeed as close as you like) would lead to an f(x) that
fell on the opposite side of y and was therefore at least
|f(c) - y| away from the first
f(x).
And so if f(c)
is not equal to y, the limit can't exist, and that means
that f(x) can't be continuous.
We started out assuming f(x) to be continuous and assuming
the existence of a missing point, y. Since we know that
functions can be continuous, it follows that if a function is
continuous over an interval, it can't have a missing point.
Comments:
If you got through all that and understood it, congratulations.
There are shorter proofs to this theorem, but they involve concepts
from a field of mathematics called topology. This proof, though
long, sticks with common concepts that we have of real numbers.
It also demonstrates the technique of trapping a real
number inside of a sequence of nested intervals. It is a powerful
method for proving theorems about the real numbers, and is used
in many other proofs.
Return to Main Text
email me at hahn@netsrq.com