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Let $X_0\in\mathbb R^n$ and $r\gt 0$. Prove that for { x in $\mathbb R^n| \parallel X - X_0 \parallel \le r$} is complete.

My thought process, since the domain is closed and bounded, the set is compact by definition. Therefore for all points in the domain, there exists a sequence in the domain that converges to the point. I am having trouble incorporating how this relates to cauchy sequences.

Any help would be appreciated, Thanks

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2 Answers 2

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Boundedness isn't necessary. Let $F$ be a closed subset of $\mathbb{R}^d$. Fix some Cauchy sequence $\{x_n\}_{n=1}^{\infty}$ in $F$. Then $\{x_n\}_{n=1}^{\infty}$ is Cauchy in $\mathbb{R}^d$, and hence it converges to some point $x$. Then $x \in F$, because $F$ is closed. So every Cauchy sequence in $F$ converges in $F$, and so $F$ is complete.

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multiple part question forgive my noobness, can we assume x will be in f wlog, what is the theorem or lemma behind the notion that every cauchy sequence in f converges in f, is is possible for a sequence with elements in f to converge to a point outside f and f still be closed? –  user5208 Jan 16 '13 at 3:29
    
In order to show that a space is complete, you need to show that every Cauchy sequence in that space converges. That's the definition. d-dimensional Euclidean space is complete, so any Cauchy sequence in $\mathbb{R}^d$ converges to some point. A set F is closed iff whenever $\{x_n\}$ is a convergent sequence in F, it converges to some point of F. I started with a Cauchy sequence in F, showed that it converges in the larger space Rd, and then concluded that the limit point lies in F since F is closed. –  anonymous Jan 16 '13 at 3:34
    
The following example might help: consider the sequence $x_n = \frac{1}{n}$ in $(0,1)$. This sequence is as Cauchy as can be and wants with all of its little heart and soul to converge to 0. But it doesn't converge to 0 in $(0,1)$, because 0 isn't in this set. So there's some Cauchy sequence which doesn't converge in $(0,1)$, so this isn't a complete space. However, it does converge in the larger space $[0,1]$, which so happens to be a complete space. –  anonymous Jan 16 '13 at 3:38
    
Ohh I see, Thanks for all the effort for that response, I appreciate it good sir/mam –  user5208 Jan 16 '13 at 3:44
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(1) What you have there isn't the precise definition of compactness. (I'm actually unsure if you're using limit point or sequential compactness, but either way it's not precise as stated.)

(2) Apply sequential compactness to an arbitrary Cauchy sequence in $X_0$. What can we say about Cauchy sequences with a convergent subsequence?

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I could be wrong, and I aften am, but but doesnt the Bolzano–Weierstrass theorem give an equivalent condition for sequential compactness in the euclidean space such that compactness is defined iff a set closed and bounded. –  user5208 Jan 16 '13 at 3:40
    
I'm not referring to that. I'm referring to "for all points in the domain, there exists a sequence in the domain that converges to the point." This is a trivial statement if you think about how this is quantified. Compare this to the precise definition of limit point/sequentially compact. –  kigen Jan 16 '13 at 3:49
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