Showing that there's a Cauchy sequence where Xn<X for all n?

In summary, to show that there exists a Cauchy sequence of rationals X1, X2,... representing X such that Xn < x for all n, we can use the density of rationals to pick a convergent sequence that converges to x from "below". This sequence can be formed by picking a rational number in (x-1,x) as the first element, then picking the next in (x-(1/2),x) as the second, and so on. By showing that this sequence is convergent and converges to x, we can conclude that it is a Cauchy sequence of rationals representing X, satisfying the given condition.
  • #1
hb1547
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Homework Statement


"If x is a real number, show that there exists a Cauchy sequence of rationals Xl, X2,... representing X such that X n < x for all n."

Homework Equations


- All Cauchy sequences are convergent
- All Cauchy sequences are bounded.

The Attempt at a Solution


These proofs that involve Cauchy sequences have been rough on me, and I'm trying to start working through them rather than just hunting for solutions.

But I just don't know quite where I should start. What should I be assuming or trying to show? Do I start with a series where Xn<X for all n that converges to x and show it's Cauchy? Just kinda stumped so far :/
 
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  • #2
Use the density of rationals and then find a convergent sequence that converges to x from "below".
 
  • #3
So should I start by assuming something is Cauchy, or by taking a sequence and then proving it to be Cauchy?
 
  • #4
No, that is not what I said. :-)

Pick a rational number in (x-1,x) call it the first element.
Pick the next in (x-(1/2),x) call it the second. Keep doing this ...
Pick the nth term from (x- (1/n) ,x).

Show the above sequence is convergent and it converges to x.

Since x_n is convergent is it cauchy ?
 
  • #5
If x is any real number, then it can be written in a decimal expansion. let xn be the number formed from the first n decimal places of x.


For example, if [itex]x= \pi[/itex], then x= 3.1415926... so the "Cauchy sequence of rational numbers" is 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, 3.1515926, ...
 
  • #6
╔(σ_σ)╝ said:
No, that is not what I said. :-)

Pick a rational number in (x-1,x) call it the first element.
Pick the next in (x-(1/2),x) call it the second. Keep doing this ...
Pick the nth term from (x- (1/n) ,x).

Show the above sequence is convergent and it converges to x.

Since x_n is convergent is it cauchy ?

Awesome, thanks! I was able to figure it out with this. Much appreciated.
 
  • #7
Any real number has an expression as a decimal expansion: [itex]a.b_1b_2b_3\cdot\cdot\cdot[/itex] or, equivalently, [itex]a+ b_1/10+ b_2/100+ b_3/1000+ \cdot\cdot\cdot[/itex], where a is an integer and every b is an integer from 0 to 9.

Look at the sequence [itex]a[/itex], [itex]a.b_1[/itex], [itex]a.b_1b_2[/itex], [itex]a.b_1b_2b_3[/itex], ...

Or, equivalently, [itex]a[/itex], [itex]a+ b_1/10[/itex], [itex]a+ b_1/10+ b_2/100[/itex], [itex]a+ b_1/10+ b_2/100+ b_3/1000[/itex], ...
 

Related to Showing that there's a Cauchy sequence where Xn<X for all n?

1. What is a Cauchy sequence?

A Cauchy sequence is a sequence of numbers where the terms eventually get closer and closer to each other, meaning that the difference between any two terms in the sequence can be made arbitrarily small by choosing a large enough index. In other words, the terms in a Cauchy sequence become increasingly more precise as the sequence progresses.

2. How do you prove that a sequence is Cauchy?

In order to prove that a sequence is Cauchy, you must show that the sequence satisfies the Cauchy criterion, which states that for any arbitrarily small positive number ε, there exists a positive integer N such that for all indices m, n >= N, the absolute value of the difference between the mth and nth terms in the sequence is less than ε. In other words, the terms in the sequence get closer and closer together as the index increases.

3. Can you give an example of a Cauchy sequence?

One example of a Cauchy sequence is the sequence of decimal approximations for π. This sequence starts with 3.1, 3.14, 3.141, and continues on with more and more decimal places, getting closer and closer to the true value of π as the sequence progresses.

4. How do you show that there's a Cauchy sequence where Xn < X for all n?

In order to show that there's a Cauchy sequence where Xn < X for all n, you can construct a sequence where the terms decrease in value as the index increases. For example, you can start with X = 1 and construct the sequence Xn = 1/(n+1). This sequence will have terms that are always less than 1, but get closer and closer to 0 as the index increases.

5. Why is showing the existence of a Cauchy sequence important?

Showing the existence of a Cauchy sequence is important because it is one of the ways to prove that a series converges, meaning that the limit of its terms exists. Convergence is a fundamental concept in calculus and real analysis, and many important theorems and applications rely on the existence of Cauchy sequences.

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