If a subsequence of a Cauchy sequence converges then the Cauchy sequence itself also converges
From Maths
Revision as of 11:31, 27 May 2016 by Alec (Talk | contribs) (Alec moved page If a subsequence of a Cauchy sequence converges then the sequence converges also converges to If a subsequence of a Cauchy sequence converges then the Cauchy sequence itself also converges without leaving a redirect: Double conv...)
Statement
Given a metric space [ilmath](X,d)[/ilmath]:
- If a Cauchy sequence [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] has a sub-sequence that is convergent then the Cauchy sequence [ilmath](x_n)_{n=1}^\infty[/ilmath] converges to the same point^{[1]}.
Proof
- Let [ilmath](x_n)_{n=1}^\infty[/ilmath] be a Cauchy sequence with a convergent subsequence [ilmath](x_{k_n})_{n=1}^\infty\longrightarrow x\in X[/ilmath]
- We wish to show that [ilmath](x_n)_{n=1}^\infty[/ilmath] converges (to [ilmath]x[/ilmath]), this means we must show:
- [ilmath]\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon][/ilmath]
- Note that by the triangle inequality of [ilmath]d[/ilmath] you should already be thinking [ilmath]d(x_n,x)\le d(x_n,x_\text{other})+d(x_\text{other},x)[/ilmath]
- [ilmath]\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon][/ilmath]
- Let [ilmath]\epsilon>0[/ilmath] be given.
- As [ilmath](x_n)_{n=1}^\infty[/ilmath] is Cauchy:
- [ilmath]\exists N_1\in\mathbb{N}\forall n_1,m_1[n_1,m_1>N\implies d(x_{m_1},x_{n_1})<\frac{\epsilon}{2}][/ilmath]
- (As it is Cauchy we know that for whatever number [ilmath]>0[/ilmath] we like there exists an $N$, here we pick the number [ilmath]\frac{\epsilon}{2} [/ilmath])
- [ilmath]\exists N_1\in\mathbb{N}\forall n_1,m_1[n_1,m_1>N\implies d(x_{m_1},x_{n_1})<\frac{\epsilon}{2}][/ilmath]
- As [ilmath](x_{k_n})_{n=1}^\infty[/ilmath] converges to [ilmath]x\in X[/ilmath]:
- [ilmath]\exists N_2\in\mathbb{N}\forall n_2\in\mathbb{N}[n_2>N_2\implies d(x_{k_{n_2} },x)<\frac{\epsilon}{2}][/ilmath]
- Choose [ilmath]N:=\text{Max}(N_1,k_{N_2})[/ilmath] (so the statements [ilmath]d(x_{k_{n_2} },x)<\epsilon[/ilmath] and [ilmath]d(x_{m_1},x_{n_1})<\epsilon[/ilmath] hold)
- Let [ilmath]n\in\mathbb{N} [/ilmath] with [ilmath]n>N[/ilmath] be given
- [ilmath]d(x_n,x)=d(x,x_n)\le d(x,x_{k_n})+\underbrace{d(x_{k_n},x_n)}_{\text{as }n\le k_n}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon[/ilmath]
- Note: Here we use that for a subsequence [ilmath]n\le k_n[/ilmath]
- [ilmath]d(x_n,x)=d(x,x_n)\le d(x,x_{k_n})+\underbrace{d(x_{k_n},x_n)}_{\text{as }n\le k_n}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon[/ilmath]
- Let [ilmath]n\in\mathbb{N} [/ilmath] with [ilmath]n>N[/ilmath] be given
- This completes the proof
- As [ilmath](x_n)_{n=1}^\infty[/ilmath] is Cauchy:
- We wish to show that [ilmath](x_n)_{n=1}^\infty[/ilmath] converges (to [ilmath]x[/ilmath]), this means we must show:
TODO: Create and link to the theorem that [ilmath]n\le k_n[/ilmath] for a subsequence.
References
Categories:
- Todo
- Theorems
- Theorems, lemmas and corollaries
- Metric Space Theorems
- Metric Space Theorems, lemmas and corollaries
- Metric Space
- Real Analysis Theorems
- Real Analysis Theorems, lemmas and corollaries
- Real Analysis
- Topology Theorems
- Topology Theorems, lemmas and corollaries
- Topology
- Functional Analysis Theorems
- Functional Analysis Theorems, lemmas and corollaries
- Functional Analysis
- First-year friendly