Can Cauchy Sequences be Bounded? Theorem 1.4 in Introduction to Analysis

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Homework Statement


Theorem 1.4: Show that every Cauchy sequence is bounded.


Homework Equations


Theorem 1.2: If a_n is a convergent sequence, then a_n is bounded.
Theorem 1.3: a_n is a Cauchy sequence \iff a_n is a convergent sequence.


The Attempt at a Solution


By Theorem 1.3, a Cauchy sequence, a_n, is a convergent sequence. By Theorem 1.2, a converging sequence must be bounded. Therefore, every Cauchy sequence is bounded.


I was just flipping through the textbook that my Analysis class will be using, "Introduction to Analysis" by Edward D. Gaughan, reading through Chapter 1. I noticed this theorem was left to an exercise, but I thought it was a bit too obvious of an answer as these two theorems in the Relevant equations were proven just before it. Is this really as simple as that?
 
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If you are allowed to use the theorems, then what you have suffices. However, textbooks are usually looking for you to prove without the given theorem, or prove the theorem itself. I would recommend trying to prove that all Cauchy sequences are convergent. Then you can say since all Cauchy sequences are convergent, all Cauchy sequences are bounded. That sounds like more fun, now doesn't it?
 
The problem does let you assume you are working in the reals (a complete metric space), right? If so, then yes, by all means use those theorems.

If you can't assume completeness, then you can't assume Cauchy sequences converge.
 
If you're not allowed to use completeness of the underlying space, then I would suggest you read the proof of "convergent sequences are bounded" and try to adjust that.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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