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Suppose $X$ is a metric space, $z$ is in $X$ and $(x_n)$ is a sequence in $X$.

Then what does it mean to say that, $z$ is in the "closure of every tail of $(x_n)$."

What does "closure" of every tail, mean ?

jim
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johny
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2 Answers2

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First, if $m\in\Bbb N$, the $m$-tail of the sequence is the subsequence $\langle x_n:n\ge m\rangle$ consisting of all of the terms from $x_m$ on. Now instead of viewing this as a sequence, consider just the set of terms in the $m$-tail: that set is $T_m=\{x_n:n\ge m\}$. It’s just a subset of $X$, so it has a closure, $$\operatorname{cl}T_m=\operatorname{cl}\{x_n:n\ge m\}\;.$$

A point $x\in X$ is therefore in the closure of every tail of $\langle x_n:n\in\Bbb N\rangle$ if

$$x\in\bigcap_{m\in\Bbb N}\operatorname{cl}T_m=\bigcap_{m\in\Bbb N}\operatorname{cl}\{x_n:n\ge m\}\;.$$

This means that if you throw away any finite number of terms at the beginning of the sequence, $x$ is still in the closure of the set of remaining terms. This will be the case if there is a subsequence of $\langle x_n:n\in\Bbb N\rangle$ converging to $x$.

Brian M. Scott
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  • I follow your answer,but, like for any subset,S of X we define the closure to be the set S, union the boundary points of S, what i still can't figure out is that, will this sequence also have some "boundary points" ? I know this is a strange question but, i am just having some difficulty in trying to imagine this for a sequence.I do understand that the sequence too, is just a subset of X, and so will have a closure but the boundary points are troubling me. – johny Oct 02 '13 at 12:28
  • @johny: Consider the sequence $\langle x_n:n\in\Bbb Z^+\rangle$ defined by $x_n=\frac1n$ when $n$ is odd and $x_n=2+\frac1n$ when $n$ is even. The $5$-tail, say, is $$\left\langle\frac15,2+\frac16,\frac17,2+\frac18,\ldots\right\rangle;,$$ so $$T_5=\left{\frac15,2+\frac16,\frac17,2+\frac18,\ldots\right};,$$ and $\operatorname{cl}T_5=T_5\cup{0,2}$: both $0$ and $2$ are accumulation points of $T_5$, and they’re its only accumulation points. – Brian M. Scott Oct 02 '13 at 12:32
  • Finally, it makes sense to me !thank you. – johny Oct 02 '13 at 12:38
  • @johny: You’re welcome. – Brian M. Scott Oct 02 '13 at 12:44
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It means that, for any $n \in \mathbb{N}$, consider the set $$ S_n = \{x_n, x_{n+1}, x_{n+2}, \ldots\} $$ Then, $z\in \overline{S_n}$ for all $n$ (here, $S_n$ is a tail of the sequence)

Prahlad Vaidyanathan
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