Consider the following proposed proof that if f: (0, 1) → R is uniformly continuous, then lim f(x) exists [special case of Proposition 3.4.6]. 2-0 1. Let {n} be a sequence in (0, 1) that converges to 0. In particular, {n} is a Cauchy sequence, so Lemma 3.4.5 implies that the image sequence {f(x)} is a Cauchy sequence, which by Theorem 2.4.5 converges to some limit in R. 2. If {n} is another sequence in (0, 1) that converges to 0, then the same reasoning shows that the sequence {f(yn)} potentially different from the limit of {f(x)}=1. converges to some limit, 3. Define a sequence {n}1 by setting 22k-1 equal to x and %2% equal to y for each natural number k. This sequence converges to 0, so the same reasoning as above implies that {f(zn)}1 converges. 4. By Proposition 2.1.17, all subsequences of a convergent sequence converge to the same limit. Since {f(x)}1 and {f(yn)}1 are subsequences of the convergent sequence {f(zn)}1, the sequence {f(yn)}a_1 converges to the same limit as does {f(x)}x=1. 5. Thus there exists a number L such that for every sequence {n} in (0, 1) converging to 0, the image sequence {f(an)}1 converges to L. So lim f(x) exists (and equals L) by Lemma 3.1.7. x-0 Which one of the following statements best describes this proposed proof? Step 4 is faulty. The proof is valid. Step 1 is faulty. Step 5 is faulty. Step 3 is faulty. Step 2 is faulty.