## Chapter 4: Continuity and the Fundamental Theorem of Algebra## 4.1 The Overall Strategy for Proving the Fundamental Theorem The difference between real numbers and arbitrary ordered fields is that
the field of real numbers was required to contain all the infinite
decimals. What this meant was that, for every infinite decimal
there had to be a real number a such that the
all the numbers of the sequence
where
are as close to the number
a as we like provided that j is chosen sufficiently large. The mathematical
parlance describes this by saying that the We will also need to understand polynomials better. Recall that a
polynomial with coefficients in a field was defined to be a formal
expression of the form
where the
are in F. We usually wrote a polynomial in the form
p(x) and substituted values b in F in for the variable to get elements
. So, the polynomial
corresponds to a set of pairs (b, p(b)) for all b in F. When we do this,
we are thinking of the polynomial as a We will be dealing with polynomials with complex coefficients. As we
have just seen, such polynomials can be thought of as defining a function from
the complex numbers to the complex numbers. But every complex number is
of the form
where a and b are real numbers. So we have another
function consisting of all pairs ((a, b), p(a + bi)) for all real numbers
pairs (a, b). Thus, the polynomial can be thought to define a function from
the set of pairs of real numbers (the In the case of a function f(x) from the real numbers to the real numbers,
we can represent the function by its In the case of a polynomial with complex coefficients, we have seen that it defines a function of two real variables with complex values. So, to each pair of real numbers (a, b), we have a complex number p(a + bi). Complex numbers are represented as pairs of reals and so we see that our complex polynomial corresponds to a four dimensional picture -- for each pair of reals, we have a pair of real numbers. Because of the difficulty of visualizing four dimensional objects, this is much less useful. Since the complex number is really a pair of numbers, we can consider the polynomial to represent a pair of functions. If p(x, y) = p(x + iy) = r + is, then we have the function consisting of the pairs ((x, y), r) and another function consisting of the pairs ((x, y), s). So geometrically, the polynomial corresponds to two surfaces. For example, if the polynomial were , then the complex function would consist of the pairs and since and , we see that the polynomial corresponds to the two functions with pairs ((x,y), x^2 + y^2) and with ((x, y), 2xy). Complex numbers z are also represented as the pair consisting of their length |z| and their argument . So, we can also think of the polynomial p with complex coefficients as two surfaces, one recording the lengths: and the other recording the argument of the functional value: . This is the approach we will take in proving the fundamental theorem of algebra. That result simply says that the length surface of a non-zero polynomial has to touch the xy-plane in at least one point. Now, the surface has no points which lie below the xy-plane because lengths are never negative. So, if we were to look for points where the surface touched the xy-plane, we should look for lowest points on the surface. We will show: - There are lowest points on the surface.
- If we are at a point P on the surface such that all nearby points are no lower than P, then P actually is on the xy-plane.
continuity,
which can be described intuitively as saying that the surface has no
shears. So, as preliminaries, we will introduce the notion of continuity
and its closely associated notion of limit. Then we will show some fundamental
properties of continuous functions. Finally, we will use these properties to
complete the proof of the Fundamental Theorem of Algebra.
## 4.2 ContinuityIntuitively, a function f(x) is said to continuous at b if functional values f(x) are as close as we would like to f(b) as soon as x is sufficiently close to b and at a place where f(x) is defined. For example, the top surface of a table defines a flat surface which is continuous. It is even continuous at the edge of the table because f(x) is not defined for points x beyond the edge of the table. On the other hand, we have a room containing only a table and consider the function consisting of the table top and the part of the floor which is not under the table, then this function is still continuous at all points except at the points corresponding to the edge of the table. Given such a point, there are points arbitrarily close to it where the functional value is defined by the table top and other points arbitrarily close to it where the functional value is defined by the floor level.
This function is continuous at all x except for x = 0. All the functional values for x near 0 are close to 0 whereas f(0) = 1. ii. A less extreme case is Again, this function is continuous at all x except x = 0. iii. This function is not continuous anywhere:
iv. This monstrosity is continuous everywhere except at (0, 0):
Let's give a more careful definition: Let f(x) be a real valued function defined for certain real values x. If f is defined at b, we want to say that f is continuous at b if f(x) is as close as we wish to f(b) for all x sufficiently close to b where f is defined. The problem is to make sense of expressions such as "as close as we wish". One might wish for different degrees of closeness at various times. To cover the most stringent case, we will interpret this as meaning the distance between the two is less than any specified positive number. So, "f(x) is as close as we wish to f(b)" means that, if you are given a maximal distance , then one must have . The expression, "for all x sufficiently close to b" means that there is a positive number such that the assertion holds for all x satisfying . So, our careful definition is:
Showing that a function is continuous can be a lot of work:
ii. Now let's try and show that the same function is continuous at x = 1. For a given , we need to have for all x with , where is yet to be chosen. Now, . So, if we want this to be small by making |x - 1| small, we can do it provided that |x + 1| is not made large in the process. But if . So, if , we would have where the last inequality holds provided we choose . This is precisely what we want provided that we also have . So, given , let be any number smaller than both 1 and . Then, if , we have because This proves that is continuous at x = 1. iii. Now, let's try to show that is continuous at all real number b. Repeating the same kind of reasoning, we would want Given an \epsilon > 0, we could choose so that it is less than 1 and Then, we would have for all x with . We will need to study the continuity of real valued functions of two variables. The careful definition is almost the same as in the one variable case:
All we really did was replace the absolute value with the distance function: the condition that is replaced with the distance from (x,y) to (a,b) be less than . We could have written Definition 1 in the same way, since absolute value gives the distance between two real numbers.
for all x in the domain of f with . But this holds for any choice of as long as it is a positive number. ii. Any linear function f(x,y) = ax + by + c is continuous at all points (x, y) in its domain. In fact, let and (r, s) be in the domain of r. We need for all (x, y) in the domain with . For (x, y) with , one has: because the square root function is increasing. Similarly . But then, we have: If we choose our so that , then the right hand side will be smaller than as desired. ## 4.3 More Continuous Functions Proving that a function is continuous using only the definition can
be quite tedious. So, we will need to develop some results which make it
easy to check that certain functions are continuous. Throughout this
section we will be dealing with real valued functions of one or two variables.
We will take their values at Given two functions f and g, we can define sums, differences, products, and quotient functions by: - (f + g)(z) = f(z) + g(z)
- (f - g)(z) = f(z) - g(z)
- (fg)(z) = f(z)g(z)
- (f/g)(z) = f(z)/g(z)
- f + g and f - g are continuous at z.
- fg is continuous at z.
- If , then f/g is continuous at z.
If is any positive number, then we can use the continuity of f and g at z to know that there is a and such that and for all points w in the domains of f and g such that (for f) and (for g). Choosing to be any number smaller than both and gives as needed. You should check through the same proof using subtraction instead of addition of functions. For products, one has Let . Choose smaller than 1 and subject to another condition which we will specify below. By the continuity of f at z, we know that there is a such that for all w in the domain of f with . Similarly, by the continuity of g at z, we know that there is a such that for all w in the domain of g with . Choose any smaller than both and . Continuing our series of inequalities, we get for all w in the domains of both f and of g with that where the last inequality follows because we add the condition to the list of conditions which is required to satisfy. Finally, let's consider the case of quotients. We need to assume that z is in the domains of f and of g as well as . The inequality looks like
Now the numerator is like the one for products and the same sort of argument will be able to handle it. The new ingredient is the denominator. In order to get an upper bound on a quotient |a|/|b|, you need to either make the numerator larger or the denominator smaller. So, we need a lower bound on |g(w)|. But and so we can choose a such that for all w in the domain of g with , one has . Since g(w) is close to g(z), it cannot be too close to zero; more specifically, where we have used:
where the last inequality would be true if we chose so that .
For assertion iii, one needs
ii. The square root function is continuous at all non-negative reals.
ii. Let a > 0 and . One has for with where is a quantity to be determined: So, if we choose so that , then one has . Now consider the case where a = 0. If , choose smaller than 1 and . If x is non-negative and , then since the square root function is increasing, and so, the square root function is even continuous at zero. ## 4.4 Theorems of Bolzano and WeierstrassThe typical discontinuity is a point where the function makes a jump. Continuous functions cannot do this:
Start with . We know that f(a) and f(b) have different signs and we are looking for an root of f. Consider . If f(d) = 0, then we have found a root. Otherwise, f(d) is non-zero and so its sign is different from either that of f(a) or that of f(b). Now replace the interval [a, b] with the interval with endpoints d and the one of a or b such that the sign of f(a) or f(b) is different from that of f(d). The new interval is called and its length is half that of its predecessor . Repeating the same process, we obtain either a root of f or an inductive definition of intervals where with and such that and are both non-zero with different signs. If we found a root, we are done; so assume that we have the sequence of intervals. According to Corollary 8 of Section 2.7.3, there is a unique real number c contained in all the intervals . It must be the case that f(c) = 0. Otherwise, the continuity of f implies that there is a such that . But, if we choose k so large that , then the every element x in satisfies . This contradicts the fact that there are x in the interval such that f(x) has a sign different from that of f(c). This completes the proof.
By induction, assume that we have defined of length . Let . If contains an e with for all x in , then set ; otherwise, let . Clearly, has length . The half that was discarded does not contain an e with f(e) larger than every f(x) in the half that was kept. By Corollary 8 of section 2.7.3, we know that there is a unique real number c contained in all the intervals . This c must be the absolute maximum. If not, there is an e in [a, b] with f(e) > f(c). Since it is not in the intersection of all the , there must be a step j in which e was in the half of the interval which was discarded. But by the choice of the discarded half, there is an in with . How large can j be? Consider the set S of non-negative integers j for which there is an in with . If j is in S, then . So, we can argue with as in the last paragraph using in place of e that there is a larger value of j in the set S. So, we see that S contains arbitrarily large integers. But, f is continuous. So, there is a such that for all x in [a, b] with . By choosing a J with , we see that every element x of with satisfies and so there can be no such x with . This contradicts our assertion that the set S contain arbitrarily large integers j, and so proves the result in the case of a function of a single variable. The same sort of proof can be used to prove Weierstrass's Theorem in the case of functions of two variables. In this case, we have a rectangle which is just the cartesian product of two intervals. Instead of dealing with a nested set of intervals, we deal with rectangles. We can let and be the two midpoints. The crucial step is: - Divide the rectangle into and . Discard a half which does not contain a point p with f(p) > f(z) for all z in the other half. Suppose you keep .
- Divide the new rectangle in half the other way, i.e. the halves are and . Discard a half which does not contain a point p with f(p) > f(z) for all z in the other half.
## 4.5 The Proof of the Fundamental Theorem of Algebra The
We want to choose c so that . Since x is a positive, the factor does not affect the argument. Since arg converts products into sums, we can solve to get One can choose any non-zero c with this argument. Then the last two terms of can be written in the form where d is a positive real number. The remaining terms are in absolute value at most for some positive e. So, for all positive x small enough so that x <|a_n|r/(2e). This completes the proof.
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