Proof

We prove parts (1) and (5), and leave the remaining parts to the exercise that follows.

To see part (1), let ϵ=1.ϵ=1. Then, since ff is continuous at c,c, there exists a δ>0δ>0 such that if |y-c|<δ|y-c|<δ and y∈Sy∈S then |f(y)-f(c)|<1.|f(y)-f(c)|<1. Since |z-w|≥||z|-|w|||z-w|≥||z|-|w|| for any two complex numbers zz and ww (backwards Triangle Inequality), it then follows that ||f(y)|-|f(c)||<1,||f(y)|-|f(c)||<1, from which it follows that if |y-c|<δ|y-c|<δ then |f(y)|<|f(c)|+1.|f(y)|<|f(c)|+1. Hence, setting M=|f(c)|+1,M=|f(c)|+1, we have that if |y-c|<δ|y-c|<δ and y∈S,y∈S, then |f(y)|≤M|f(y)|≤M as desired.

To prove part (5), we first make use of part 1. Let δ1,M1δ1,M1 and δ2,M2δ2,M2 be chosen so that if |y-c|<δ1|y-c|<δ1 and y∈Sy∈S then

and if |y-c|<δ2|y-c|<δ2 and y∈Sy∈S then

Next, let ϵ'ϵ' be the positive number |g(c)|/2.|g(c)|/2. Then, there exists a δ'>0δ'>0 such that if |y-c|<δ'|y-c|<δ' and y∈Sy∈S then |g(y)-g(c)|<ϵ'=|g(c)|/2.|g(y)-g(c)|<ϵ'=|g(c)|/2. It then follows from the backwards triangle inequality that

Now, to finish the proof of part (5), let ϵ>0ϵ>0 be given. If |y-c|<min(δ1,δ2,δ')|y-c|<min(δ1,δ2,δ') and y∈S,y∈S, then from Inequalities (3.1), (3.2), and (3.3) we obtain

Finally, using the continuity of both ff and gg applied to the positive numbers ϵ1=ϵ/(4M2|g(c)|2)ϵ1=ϵ/(4M2|g(c)|2) and ϵ2=ϵ/(4M1|g(c)|2),ϵ2=ϵ/(4M1|g(c)|2), choose δ>0,δ>0, with δ<min(δ1,δ2,δ'),δ<min(δ1,δ2,δ'), and such that if |y-c|<δ|y-c|<δ and y∈Sy∈S then |f(y)-f(c)|<ϵ4M2/|g(c)|2|f(y)-f(c)|<ϵ4M2/|g(c)|2 and |g(c)-g(y)|<ϵ4M1/|g(c)|2.|g(c)-g(y)|<ϵ4M1/|g(c)|2. Then, if |y-c|<δ|y-c|<δ and y∈Sy∈S we have that

as desired.

Proof

Let ϵ>0ϵ>0 be given. Because gg is continuous at the point f(c),f(c), there exists an α>0α>0 such that |g(t)-g(f(c))|<ϵ|g(t)-g(f(c))|<ϵ if |t-f(c)|<α.|t-f(c)|<α. Now, using this positive number α,α, and using the fact that ff is continuous at the point c,c, there exists a δ>0δ>0 so that |f(s)-f(c)|<α|f(s)-f(c)|<α if |s-c|<δ.|s-c|<δ. Therefore, if |s-c|<δ,|s-c|<δ, then |f(s)-f(c)|<α,|f(s)-f(c)|<α, and hence |g(f(s))-g(f(c))|=|g∘f(s)-g∘f(c)|<ϵ,|g(f(s))-g(f(c))|=|g∘f(s)-g∘f(c)|<ϵ, which completes the proof.

Proof

Suppose first that ff is continuous at c,c, and let {xn}{xn} be a sequence of elements of SS that converges to c.c. Let ϵ>0ϵ>0 be given. We must find a natural number NN such that if n≥Nn≥N then |f(xn)-f(c)|<ϵ.|f(xn)-f(c)|<ϵ. First, choose δ>0δ>0 so that |f(y)-f(c)|<ϵ|f(y)-f(c)|<ϵ whenever y∈Sy∈S and |y-c|<δ.|y-c|<δ. Now, choose NN so that |xn-c|<δ|xn-c|<δ whenever n≥N.n≥N. Then if n≥N,n≥N, we have that |xn-c|<δ,|xn-c|<δ, whence |f(xn)-f(c)|<ϵ.|f(xn)-f(c)|<ϵ. This shows that the sequence {f(xn)}{f(xn)} converges to f(c),f(c), as desired.

We prove the converse by proving the contrapositive statement; i.e., we will show that if ff is not continuous at c,c, then there does exist a sequence {xn}{xn} that converges to cc but for which the sequence {f(xn)}{f(xn)} does not converge to f(c).f(c). Thus, suppose ff is not continuous at c.c. Then there exists an ϵ0>0ϵ0>0 such that for every δ>0δ>0 there is a y∈Sy∈S such that |y-c|<δ|y-c|<δ but |f(y)-f(c)|≥ϵ0.|f(y)-f(c)|≥ϵ0. To obtain a sequence, we apply this statement to δδ's of the form δ=1/n.δ=1/n. Hence, for every natural number nn there exists a point xn∈Sxn∈S such that |xn-c|<1/n|xn-c|<1/n but |f(xn)-f(c)|≥ϵ0.|f(xn)-f(c)|≥ϵ0. Clearly, the sequence {xn}{xn} converges to cc since |xn-c|<1/n.|xn-c|<1/n. On the other hand, the sequence {f(xn)}{f(xn)} cannot be converging to f(c),f(c), because |f(xn)-f(c)||f(xn)-f(c)| is always ≥ϵ0.≥ϵ0.

This completes the proof of the theorem.