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

  • Thread starter hb1547
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  • #1
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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 :/
 

Answers and Replies

  • #2
Use the density of rationals and then find a convergent sequence that converges to x from "below".
 
  • #3
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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
HallsofIvy
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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
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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
HallsofIvy
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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], ...
 

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