prove R^n is complete

[galactus]

Hi all. I have been studying a little advance calc. I have a question.

"Starting from the fact that R, equipped with its usual distance function is complete, prove that R^n equipped with the sum, maximum, or Euclidean norm is complete."

I am just starting to get into all these advanced calc terms like Complete and so forth. All these years I never bothered to study any advanced calc. Just what I absorbed through osmosis. I use calculus, but never studied the nuts and bolts of it in any depth.

I know that a metric space is complete if every Cauchy sequence in X converges to a limit in X.

 

[daon]

So you're only able to use the fact that \(\displaystyle R^1\) is complete? Here's an illustration for \(\displaystyle R^2\) that I think works. The case for general n will probably follow from induction.

If you know that \(\displaystyle x_k \to x, y_k \to y\),

Then, given \(\displaystyle \epsilon\), choose \(\displaystyle N_x\) and \(\displaystyle N_y\) so that \(\displaystyle |x_n-x_m|\) and \(\displaystyle |y_n-y_m|\) are less than \(\displaystyle \frac{\epsilon}{2}\) respectively, whenever \(\displaystyle m,n \ge N=\max\{N_x,N_y\}\).

Then \(\displaystyle d \left((x_n,y_n),(x_m,y_m) \right)=\sqrt{(x_n-x_m)^2 + (y_n-y_m)^2} \le \sqrt{\frac{\epsilon^2}{4} + \frac{\epsilon^2}{4}} = \frac{\epsilon}{\sqrt{2}} < \epsilon}\).

Rudin has a nice proof, but it assumes some things for compact metric spaces.

 

[daon]

Sorry, I think I just did that backwards.

Suppose \(\displaystyle \{(x_n,y_n)\}\) is cauchy,

If \(\displaystyle \epsilon>0\) we want to show that for some \(\displaystyle N \in \mathbb{N}, n,m \ge N \implies\) \(\displaystyle |x_n-x_m|<\epsilon\) and \(\displaystyle |y_n -y_m|< \epsilon\) so that \(\displaystyle x_k\) and \(\displaystyle y_k\) converge to some \(\displaystyle x, y \in R\) respectively by assumption, and therefore \(\displaystyle (x_k,y_k) \to (x,y) \in R^2\)

So we are assuming for any positive \(\displaystyle \epsilon\) that there is an \(\displaystyle N\) such that if \(\displaystyle m,n \ge N\):

\(\displaystyle 0 \le \sqrt{(x_n-x_m)^2 + (y_n-y_m)^2} < \epsilon \,\,\)

But that immediately implies:

\(\displaystyle \left (|x_n-x_m|=\sqrt{(x_n-x_m)^2} < \epsilon \right) \wedge \left(|y_n-y_m| = \sqrt{(y_n-y_m)^2} < \epsilon \right)\)

This is "obvious" since:

\(\displaystyle \sqrt{a^2 + b^2} < c \implies |a| < c\), for if \(\displaystyle |a| \ge c\) then \(\displaystyle \sqrt{a^2 + b^2} \ge \sqrt{c^2+b^2} \ge c\).

So we have that \(\displaystyle x_k\) and \(\displaystyle y_k\) are both Cauchy, and by assumption must converge to some \(\displaystyle x,y \in R\).

 

[galactus]

Thanks Daon:

That's cool. I thought it was about time I studied some analysis. I have some books and want to make use of them.

Thanks again, Daon, you're aces.

Here's one I got:

Prove \(\displaystyle \lim_{n\to {\infty}}\frac{1}{n}=0\)

Need to fins \(\displaystyle \epsilon>0\), a value N s.t. \(\displaystyle |x_{n}-x|<\epsilon, \;\ \forall n>N\)

So, \(\displaystyle |x_{n}-x|=|\frac{1}{n}-0|=|\frac{1}{n}|\)

\(\displaystyle \frac{1}{n}<\epsilon\Rightarrow \frac{1}{\epsilon}<n\)

Choose N as the least integer\(\displaystyle >\frac{1}{\epsilon}\), then \(\displaystyle \forall n>N \;\ |\frac{1}{n}|<\frac{1}{N}\leq {\epsilon}\).

I am just learning what all these terms like Completeness and Banach Space means. Cool stuff. This is real math. I like it, Though it can fry a few synapses once in a while.

Oh, BTW, have you ever seen a symbol used for 'such that' that looks like a backward epsilon?. That is what I am used to using for 'such that', though we do not see it that often. I wonder if LaTex has it and what the code would be for it.

 

[daon]

Your proof looks dandy.

Also, part of the fun (for me, at least) is NOT showing how you got your \(\displaystyle N, \delta\), etc., just pulling it out of thin air for your proof :wink:.

I've never seen nor used that symbol, but most browsers should display this character: ?. That it? Looks a bit like an "is an element of" symbol. edit: "\ni" will make \(\displaystyle \ni\). "\backepsilon" will make \(\displaystyle \backepsilon\).

What if we wrote "There exists an x in [a,b] such that x is an element of [c,d]."

\(\displaystyle \exists x \in [a,b]\ni x \in [c,d]\) :shock:

 

[galactus]

Hey Daon:

Here is another analysis problem. I was thinking perhaps the Bolsano-Weierstarss theorem may be useful. I have been reading up on the cluster points.

"Let \(\displaystyle {x_{n}}\) be a sequence of real numbers. A point \(\displaystyle c\in \Re\cup {\pm \infty}\) is called a cluster point of \(\displaystyle (x_{n})\) if there is a convergent subsequence with limit c. Let C be the set of cluster points of (x_n). Prove that C is closed and \(\displaystyle lim sup \;\ x_{n}=maxC \;\ and \;\ liminf \;\ x_{n}=minC"\)

Can we show there is a subsequence \(\displaystyle x_{{n}_{k}} \;\ of \;\ x_{n} \;\ s.t. \;\ x_{{n}_{k}}\rightarrow x\).

From what I gather, getting rid of a finite number of elements from a sequence does not change the cluster point.

By definition of infinum, that for every k there exists a cluster point such that \(\displaystyle c_{k}<x+\frac{1}{2k}\).

Just some thoughts I am throwing out there.

 

[daon]

Are you trying to show that \(\displaystyle C\) is a closed set in the standard topology on R?

Yes. But, it's no big deal. I was just studying this stuff in my spare time and thought this one was interesting.

Thanks.

EDIT: I am so sorry. I must have accidentally deleted your post when I quoted. How I managed that I do not know.

 

[daon]

Haha, its cool. I was freaked out for a second.

Any countable subset of R is closed, since the complement is a union of a countable number of open intervals.