324 CHAPTER 15. SEQUENCES, COMPACTNESS, CONTINUITY

Note how the target function is not continuous although each function in the sequenceis. The next theorem shows that this kind of loss of continuity never occurs when you haveuniform convergence. The theorem holds generally when S ⊆ X a normed linear space andf,fn have values in Y another normed linear space. You should fill in the details to be sureyou understand this. You simply replace |·| with ∥·∥ for an appropriate norm.

Theorem 15.13.3 Let fn : S →Cq be continuous and let fn converge uniformly tof on S. Then if fn is continuous at x ∈ S, it follows that f is also continuous at x.

Proof: Let ε > 0 be given. Let N be such that if n ≥ N, then

supy∈S

|fn (y)−f (y)| ≡ ||fn −f ||∞<

ε

3

Pick such an n. Then by continuity of fn at x, there exists δ > 0 such that if |y−x|< δ ,then |fn (y)−fn (x)|< ε

3 . Then if |y−x|< δ ,y ∈ S, then

|f (x)−f (y)| ≤ |f (x)−fn (x)|+ |fn (x)−fn (y)|+ |fn (y)−f (y)|

<ε

3+

ε

3+

ε

3= ε

Thus f is continuous at x as claimed.

15.14 Fundamental Theorem of AlgebraRecall from the chapter on prerequisite material the basic properties of complex numbers,complex absolute value and so forth. See Definition 1.13.3 and the following material afterthis definition. If you have a sequence of complex numbers {zk} where zk = xk + iyk then to

say that |zk| is bounded is to say that√

x2k + y2

k is bounded. In other words, the ordered pairs

(xk,yk) are in a bounded subset of R2. Also to say that limk→∞ |zk − z|= 0 is the definitionof what you mean by limk→∞ zk = z and it is the same as saying that limk→∞ (xk,yk) = (x,y)where z = x+ iy. Thus, if you have a bounded sequence of complex numbers {zk} , youmust have {xk} and {yk} both be a bounded sequence in R and so {xk} is contained in someinterval [a,b] and {yk} is contained in some interval [c,d]. Thus such a bounded sequencemust have a subsequence, still denoted as zk such that xk → x ∈ [a,b] and yk → y ∈ [c,d].This yields the following simple observation sometimes called the Weierstrass Bolzanotheorem.

Theorem 15.14.1 Let {zk} be a sequence of complex numbers such that |zk| is abounded sequence of real numbers. Then there exists a subsequence

{znk

}and a complex

number z such that limk→∞ znk = z. Sets of the form K ≡ {z ∈ C : |z| ≤ r} are sequentiallycompact.

Proof: It only remains to verify the last assertion. Letting {zk} ⊆ K, the above discus-sion shows that there exists z and a subsequence

{znk

}such that znk → z. It only remains

to verify that z ∈ K. However, this is clear from the triangle inequality. Indeed,

|z| ≤ |z− zk|+ |zk| ≤ |z− zk|+ r

Hence,|z| ≤ lim

k→∞

∣∣z− znk

∣∣+ r = r.