Difference between revisions of "Every lingering sequence has a convergent subsequence/Statement"
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(Created page with "==Statement== <onlyinclude> Let {{M|(X,d)}} be a metric space, then{{rITTGG}}: * {{MM|1=\forall(x_n)_{n=1}^\infty\subseteq X\left[\left(\exists x\in X\ \forall\epsilon>0[\...") |
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Let {{M|(X,d)}} be a [[metric space]], then{{rITTGG}}: | Let {{M|(X,d)}} be a [[metric space]], then{{rITTGG}}: | ||
* {{MM|1=\forall(x_n)_{n=1}^\infty\subseteq X\left[\left(\exists x\in X\ \forall\epsilon>0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert=\aleph_0]\right)\implies\left(\exists(k_n)_{n=1}^\infty\subseteq\mathbb{N}\left[(\forall n\in\mathbb{N}[k_n<k_{n+1}])\implies\left(\exists x'\in X\left[\lim_{n\rightarrow\infty}(x_{k_n})=x'\right]\right)\right]\right)\right]}} | * {{MM|1=\forall(x_n)_{n=1}^\infty\subseteq X\left[\left(\exists x\in X\ \forall\epsilon>0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert=\aleph_0]\right)\implies\left(\exists(k_n)_{n=1}^\infty\subseteq\mathbb{N}\left[(\forall n\in\mathbb{N}[k_n<k_{n+1}])\implies\left(\exists x'\in X\left[\lim_{n\rightarrow\infty}(x_{k_n})=x'\right]\right)\right]\right)\right]}} | ||
| − | This is just a verbose way of | + | This is just a verbose way of expressing the statement that: |
* Given a sequence {{M|1=(x_n)_{n=1}^\infty\subseteq X}} if it is a [[lingering sequence]] then it has a [[subsequence]] that [[convergence (sequence)|converges]] | * Given a sequence {{M|1=(x_n)_{n=1}^\infty\subseteq X}} if it is a [[lingering sequence]] then it has a [[subsequence]] that [[convergence (sequence)|converges]] | ||
</onlyinclude> | </onlyinclude> | ||
==References== | ==References== | ||
<references/> | <references/> | ||
Latest revision as of 19:57, 9 November 2016
Statement
Let [ilmath](X,d)[/ilmath] be a metric space, then[1]:
- [math]\forall(x_n)_{n=1}^\infty\subseteq X\left[\left(\exists x\in X\ \forall\epsilon>0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert=\aleph_0]\right)\implies\left(\exists(k_n)_{n=1}^\infty\subseteq\mathbb{N}\left[(\forall n\in\mathbb{N}[k_n<k_{n+1}])\implies\left(\exists x'\in X\left[\lim_{n\rightarrow\infty}(x_{k_n})=x'\right]\right)\right]\right)\right][/math]
This is just a verbose way of expressing the statement that:
- Given a sequence [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] if it is a lingering sequence then it has a subsequence that converges
