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How to technically refer to or describe a sequence with both a beginning and an end as well as infinitely many elements between them?

      $ a_1 ~,~ a_2 ~,~ a_3 ~,~ \ldots ~,~ a_{-3} ~,~ a_{-2} ~,~ a_{-1} $

Doesn't seem to break any generic definition of sequence that allows for doubly infinite sequences.

This is related to an open-ended logic puzzle at Puzzling StackExchange.

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    $\begingroup$ I would call that a $(\omega + \omega^*)$-sequence, see e.g. mathworld.wolfram.com/OrderType.html. But I don't think a canonical name exists. $\endgroup$ – PseudoNeo Jan 6 '16 at 10:40
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    $\begingroup$ How does an "infinite" sequence have both a beginning and an end? $\endgroup$ – user160738 Jan 6 '16 at 10:52
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    $\begingroup$ The ordinal (and the order type) $\omega + 1$ is infinite and has a beginning and an end. So does the set $0, 1/2, 2/3, ... n/(n+1), ... 1$. @PseudoNeo I'd call it an $\omega+\omega^*$ sequence too. It's convergence properties are not very interesting — such a sequence always converges, to $a_{-1}$. $\endgroup$ – BrianO Jan 6 '16 at 16:38
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    $\begingroup$ By the way, $\omega+\omega^*$ is the order type of $\{\frac1n:n\in\Bbb Z\text{ and }n\ne0\}$. That is, it's the order type of $\{-1,-\frac12,-\frac13,\dots,\frac13,\frac12,1\}$. $\endgroup$ – Akiva Weinberger Jan 6 '16 at 20:22
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    $\begingroup$ @human I'd actually use $\omega + 1$: $a_0, a_1, a_2, \dotsc, a_{\omega}$. (Logicians count from 0 :) But if I wanted to stick to $\Bbb Z$, indexes that require no explanation, I might use $a_1, a_2, \dotsc, a_0$. Of course that gets harder & harder to do as the (generalized) sequences get longer & longer, e.g. enumerating all $m + n/(n+1)$, of order type $\omega^2$, alias $\Bbb N^2$ in lexicographic order. $\endgroup$ – BrianO Jan 7 '16 at 18:51
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Originally I responded to a question in a comment, How does an "infinite" sequence have both a beginning and an end?

The ordinal (and the order type) $\omega + 1$ is infinite and has a beginning as well as an end. So does the set $0,1/2,2/3,...n/(n+1),...1$. Like @PseudoNeo, I too would call the doubly-infinite sequence of the original question an $\omega + \omega^*$ sequence. I'd be inclined to call it a "sequence" for two reasons:

  • its domain is totally ordered, and
  • the total ordering is discrete.

It's an example of the more general notion of a net in a topological space, so-called and popularized by Kelley in his book General Topology:

A net in a space $X$ is a function $u\mapsto x_u\colon (D,\preceq) \to X$ on a directed preorder $(D,\preceq)$:

  • $\preceq$ is reflexive and transitive on $D$ (it preorders $D$), and
  • $\preceq$ is directed, in the sense that for every $u, v\in D$ there is $w\in D$ such that $u\preceq w$ and $v\preceq w$.

A net $(x_u)_{u\in D}$ converges to $x\in X$ iff for every neighborhood $U$ of $x$, the net is "eventually in $U$", meaning, there is $u\in D$ such that for all $v\succeq u$, $x_v \in U$. Note that when $(D,\preceq)$ is the integers with the usual ordering (alias $\omega$), this definition of convergence is precisely the standard definition of convergence of a sequence.

When $(D,\preceq)$ has a greatest element $\overline{d}$, convergence of nets on $(D,\preceq)$ is not very interesting — such a sequence always converges to $x_{\overline{d}}$. Thus the (generalized) sequence of the original question converges to $a_{-1}$, regardless of the other elements of the sequence. It might as well be a one-element sequence.

The net analog of a subsequence is a subnet: a subnet of $(x_d)_{d\in D}$ is an increasing cofinal map $c\mapsto d_c\colon (C, \le)\to(D,\preceq)$, composed with the original net $(x_d)$ to give a new net $c\mapsto x_{d_c}$. Here, "increasing" means "not necessarily strictly", $(C,\le)$ is a directed preorder, and "cofinal" means that for all $d\in D$ there is $c\in C$ with $d\preceq d_c$. If a net converges to a point $x$, then any subnet of that net also converges to $x$.

In terms of the example, $a_1, a_2, a_3, \dotsc$ is not a subnet of $a_1, a_2, a_3, \dotsc, a_{-3}, a_{-2}, a_{-1}$, as the $\omega$ part of $\omega+\omega^*$ isn't cofinal -- it doesn't "go all the way". Notice, though, that the one-element sequence $(a_{-1})$ is a subnet of the $\omega+\omega^*$ sequence.

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    $\begingroup$ I have seen this referred to as an $(\omega,\omega^*)$ sequence but using the plus sign seems just as good if not better/ $\endgroup$ – DanielWainfleet Jan 7 '16 at 2:07
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    $\begingroup$ '+' is a standard operation on order types. I agree, it is better ;/ It goes back to Cantor! More concise, too -- and it really is associative (up to isom.) though not commutative. I haven't seen "$(\omega, \omega^*)$ used to mean that, and personally that notation makes me think of other things first ... e.g. infinitary combinatorics, or even Chang's Conjecture (though I'd realize soon enough that I was mistaken). $\endgroup$ – BrianO Jan 7 '16 at 2:28
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    $\begingroup$ $+$ and $\times$ are standard operations on order types… but exponentiation seems not to be (unless the base and exponent are ordinals). $\endgroup$ – Akiva Weinberger Jan 7 '16 at 2:41
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    $\begingroup$ In Kunen, Set Theory, there is a brief reference to the the existence of an "$(\omega_1, \omega_1^*) gap"$ in $P(\omega)_{/fin}$.But he uses <a,b> for an ordered pair. I agree that round brackets are over-used. The word "normal" is also overloaded. $\endgroup$ – DanielWainfleet Jan 7 '16 at 2:45
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    $\begingroup$ Speaking of summation and order types, I think that every countable order type can be expressed as $\displaystyle\sum_{x\in\Bbb Q}f(x)$ for some $f:\Bbb Q\to\omega_1$, though I'm not sure. I'm not sure how you'd write that conjecture with user254665's notation. EDIT: Wait, that conjecture's trivial, isn't it? We only need $f:\Bbb Q\to\{0,1\}$. $\endgroup$ – Akiva Weinberger Jan 7 '16 at 2:52

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