# Proof that two equivalent sequences are both Cauchy sequences

• yucheng
In summary, the sequence a_n-b_n is equivalent to the sequence b_n-a_n iff for each ##\epsilon>0##, there exists an N>0 such that for all n>N, ##|a_n-b_n|<\epsilon##.f

#### yucheng

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.

Delta2
There's no need to argue by contradiction.

Intuitively, you know that ##|a_n-b_n|## is small for large ##n## and also ##|a_p-a_q|## is small for large ##p,q##. You want to show that ##|b_p-b_q|## is small for large ##p,q##. Do you see why you can use the triangle inequality here? Hint: ##b_p-b_q=(b_p-a_p)+(a_p-a_q)+(a_q-b_q).##

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etotheipi, yucheng and Delta2
Do you see why you can use the triangle inequality here? Hint: ##b_p-b_q=(b_p-a_p)+(a_p-a_q)+(a_q-b_q).##

Is there a way to prove it without resorting to the triangle inequality?

I don't think so, but why would you want to avoid the triangle inequality? You're not going to get far in an analysis class without it.

Delta2 and yucheng
I don't think so, but why would you want to avoid the triangle inequality? You're not going to get far in an analysis class without it.
Oh really? I am not even acquainted with it! Thanks for your advice. I'll make sure I learn it. ;-)

Delta2
Do you see why you can use the triangle inequality here? Hint: ##b_p-b_q=(b_p-a_p)+(a_p-a_q)+(a_q-b_q).##

(Me after staring at the screen for some time) Let me try. Given ##(a_n)_{n=0}^{\infty}## and ##(b_n)_{n=0}^{\infty}## are equivalent, this implies that for some N>=0, there exists ##|a_n-b_n|\leq \epsilon /2## (Let's brush over this little assumption). Now, we see that ##b_p-b_q=(b_p-a_p)+(a_p-a_q)+(a_q-b_q)\leq |b_p-a_p|+|a_p-a_q|+|a_q-b_q| \leq \epsilon /2+\delta+\epsilon /2 = \epsilon +\delta## by the triangle inequality, thus we can rewrite the RHS as ##RHS \equiv \epsilon##, by the definition of ##\epsilon##, we conclude that ##b_p-b_q \leq \epsilon##.

I hope this is complete :)

Delta2
Note quite. You haven't said what your ##\delta## is or how it relates to ##\varepsilon##; it looks like you are saying ##\varepsilon+\delta=\varepsilon##? What you want to prove is just: for any ##\varepsilon>0##, there exists ##N\in\mathbb{N}## such that ##|b_q-b_p|<\varepsilon## when ##p,q>N##. It seems to me that the simplest way to do this would be to make each of the three terms in the above sum smaller than ##\varepsilon/3.##

Delta2 and yucheng
The main idea is that you have two sequences. They are equivalent. We want to show that if one of them is Cauchy, so is the other. What does it mean intuitively for a sequence to be Cauchy? It means that somewhere after a large enough N (a term in the sequence) ANY TWO points of the sequence will be very very close to each other ( if we want the points to be closer, we can just move the N further along).

So if we first assume (an) is Cauchy, WTS (bn) is Cauchy. Use Infrared suggestion in post 7.

Look at your post #6. What do you know about |ap- aq|? Hint. It is your assumption.

What do you know about |bp-ap|? it is also your assumption. What else is missing?

I think a better question to ask you. Do you understand what the definition of Cauchy is?

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yucheng and Delta2
Intuitively is easy. If they are equivalent it means that the terms of one are very close to the respective terms of the other. So if one is Cauchy and we have ##|a_n-a_m|<\varepsilon## then we will also have ##|b_n-b_m|<\varepsilon## because ##b_n## is close to ##a_n## and ##b_m## is close to ##a_m## so kinda of $$|b_n-b_m|\approx |a_n-a_m|<\varepsilon$$.
But of course the neatest way to prove the above is by triangular inequality as post#2 suggests very elegantly.

yucheng