Real analysis (cauchy sequence)

If it is true, give a simple reason to support it: if it is false, provide
a counterexample.
(a) Let a function f: (0, 1)-->R be continuous. If (x_n) is a cauchy sequence
in (0, 1), then (f(x_n)) is also a cauchy sequence.
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For first, may someone explain what is (x_n) is a
cauchy sequence in (0, 1) ?

"(x_n) is a cauchy sequence in (0, 1)" means that (x_n) is a cauchy sequence, all of whose entries and limit are in the interval (0, 1).

That is, given ε > 0, there exists a positive integer N such that for all m, n > N, we have
|x_m - x_n| < ε for all m, n > N (and as above, x_k is in (0, 1) for all k).
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(a) This is false; consider the sequence {1/n : n is in N} which is a subset of (0, 1), along with the function f(x) = 1/x. Then, {f(1/n) = n : n in N} is not Cauchy (since it diverges to infinity).

I hope this helps!

(x_n) is a Cauchy sequence in (0,1) if 0 < x_n < 1 for each n and, given e>0, there exists a positive integer N such that if n, m > N, |x_n - x_m| < e.

For (a) we can give a counterexample. Let f(x) = 1/x and x_n = 1/n, n = 1, 2, .... Then (x_n) is a Cauchy sequence in (0,1). For given e > 0, let N be any integer greater than 2/e. If n > m > N, |x_n - x_m| = |1/n - 1/m| ≤ 1/n + 1/m < 2/m < e. But f(x_n) = x_n is not Cauchy because it is unbounded.