- #1

yucheng

- 232

- 57

- Homework Statement
- Given ##(a_n)_{n=1}^\infty## and ##(b_n)_{n=1}^\infty## are equivalent, proof that ##(a_n)_{n=1}^\infty## is a Cauchy sequence iff ##(b_n)_{n=1}^\infty## is a Cauchy sequence. (Tao's Analysis 1, Exercise 5.2.1)

- Relevant Equations
- N/A

Let us just lay down some definitions. Both sequences are equivalent iff for each ##\epsilon>0## , there exists an N>0 such that for all n>N, ##|a_n-b_n|<\epsilon##.

A sequence is a Cauchy sequence iff ##\forall\epsilon>0:(\exists N>0: (\forall j,k>N:|a_j-a_k|>\epsilon))##.

We proceeded by contradiction. Suppose ##(a_n)_{n=1}^\infty## is not a Cauchy sequence, this means that there exists ##\epsilon## such that at least one pair ##|a_j-a_k|>\epsilon##, ##\forall N>0 \land j,k>N##.

Thus, let us choose an ##\epsilon## that fulfils the above statement.

Since both sequences are equivalent, ##|a_n-b_n|<\epsilon##.

Since ##(b_n)_{n=1}^\infty## is a Cauchy sequence, we can find a N by definition that fulfils the ##\epsilon## we have chosen.

With this N, we choose a ##|a_n-a_k|>\epsilon##, where n,k > N, from sequence ##(a_n)_{n=1}^\infty##.

From ##(b_n)_{n=1}^\infty## , we also choose ##|b_n-b_k|<\epsilon##, n,k > N.

We arrive at ##|b_n-b_k|<|a_n-a_k|##. However, I am now officially stuck. Thanks in advance.

A sequence is a Cauchy sequence iff ##\forall\epsilon>0:(\exists N>0: (\forall j,k>N:|a_j-a_k|>\epsilon))##.

We proceeded by contradiction. Suppose ##(a_n)_{n=1}^\infty## is not a Cauchy sequence, this means that there exists ##\epsilon## such that at least one pair ##|a_j-a_k|>\epsilon##, ##\forall N>0 \land j,k>N##.

Thus, let us choose an ##\epsilon## that fulfils the above statement.

Since both sequences are equivalent, ##|a_n-b_n|<\epsilon##.

Since ##(b_n)_{n=1}^\infty## is a Cauchy sequence, we can find a N by definition that fulfils the ##\epsilon## we have chosen.

With this N, we choose a ##|a_n-a_k|>\epsilon##, where n,k > N, from sequence ##(a_n)_{n=1}^\infty##.

From ##(b_n)_{n=1}^\infty## , we also choose ##|b_n-b_k|<\epsilon##, n,k > N.

We arrive at ##|b_n-b_k|<|a_n-a_k|##. However, I am now officially stuck. Thanks in advance.