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Complete metric space

Ubistvo

New member
Dec 19, 2012
10
Let $(M,d)$ be a metric space and let $X_n$ be a sequence such that $X_{n+1}\subset X_n.$ Prove that if $\text{diam}(X_n)\to0$ as $x\to\infty$ and $\bigcap_{n=1}^\infty X_n\ne\varnothing,$ then $(M,d)$ is complete.

Let $x_n$ be a Cauchy sequence and consider $X_n=\overline{\{x_m:m\ge n\}},$ then $X_n$ is closed and $X_{n+1}\subset X_n,$ but I don't see how to prove that $\text{diam}(X_n)\to0,$ any ideas?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,716
Let $(M,d)$ be a metric space and let $X_n$ be a sequence such that $X_{n+1}\subset X_n.$ Prove that if $\text{diam}(X_n)\to0$ as $x\to\infty$ and $\bigcap_{n=1}^\infty X_n\ne\varnothing,$ then $(M,d)$ is complete.

Let $x_n$ be a Cauchy sequence and consider $X_n=\overline{\{x_m:m\ge n\}},$ then $X_n$ is closed and $X_{n+1}\subset X_n,$ but I don't see how to prove that $\text{diam}(X_n)\to0,$ any ideas?
You have made the right choice for $X_n$. Now use the fact that $\{x_n\}$ is a Cauchy sequence to conclude that $d(x_m,x_n)$ can be made small for all $m\geqslant n$ and hence that $\text{diam}(X_n)$ can be made small, provided that $n$ is suitably large.
 

Ubistvo

New member
Dec 19, 2012
10
Okay, so, since $x_n$ is Cauchy, then exists $N\in\mathbb N$ such that for $m,n\ge N$ we have $d(x_m,x_n)<\epsilon,$ but I don't see how to relate $\text{diam}(X_n)$ with the Cauchy sequence. I know that $\text{diam}(X_n)$ is a supremum, so by definition, this supremum is greater than a distance.

I have the ideas but I don't see how to write them analytically. :(
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,716
Okay, so, since $x_n$ is Cauchy, then exists $N\in\mathbb N$ such that for $m,n\ge N$ we have $d(x_m,x_n)<\epsilon,$ but I don't see how to relate $\text{diam}(X_n)$ with the Cauchy sequence. I know that $\text{diam}(X_n)$ is a supremum, so by definition, this supremum is greater than a distance.

I have the ideas but I don't see how to write them analytically. :(
If $x_m$ and $x_k$ are in $X_n$ (so that $m\geqslant n$ and $k\geqslant n$), then $d(x_m,x_n)<\varepsilon$ and $d(x_k,x_n)<\varepsilon$. By the triangle inequality, $d(x_m,x_k)<2\varepsilon$. Since elements of that form are dense in $X_n$, it follows that the distance between any two points in $X_n$ is at most $2\varepsilon$. Thus $\text{diam}(X_n)\leqslant 2\varepsilon$.
 

Ubistvo

New member
Dec 19, 2012
10
Okay, does the following argument work?

Since $d(x_m,x_n)<\epsilon$ for all $m,n\ge N,$ then $\sup\{d(x_n,x_m):x_n,x_m\in X_N\}\le \epsilon,$ so this implies that $\text{diam}(X_N)<\epsilon,$ but does actually follow that $\text{diam}(X_n)<\epsilon$?

I'm sorry, I got confused when you used $k.$
Thanks for the help!

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Okay so it does, since $n\ge N$ we have that $X_n\subset X_N$ then $\text{diam}(X_n)<\text{diam}(X_N)<\epsilon,$ so this proves that $\text{diam}(X_n)\to0,$ then now it follows that $\bigcap_{n=1}^\infty X_n\ne\varnothing,$ so let $x\in \bigcap_{n=1}^\infty X_n,$ then we need to show that $x_n\to x.$ Since $\text{diam}(X_n)\to0,$ there exists $N\in\mathbb N$ so that $\text{diam}(X_n)<\dfrac\epsilon2,$ since $x\in X_N$ then there is $m\ge N$ such that $d(x,x_m)<\dfrac\epsilon2,$ so for each $n\ge N$ we have $d(x,x_n)\le d(x,x_m)+d(x_m,x_n)\le\dfrac\epsilon2+\text{diam}(X_N)\le\dfrac\epsilon2+\dfrac\epsilon2<\epsilon,$ so $d(x,x_n)<\epsilon$ which implies that $x_n\to x,$ and then $(M,d)$ is complete.

Is it correct?
 
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