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.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ How does an "infinite" sequence have both a beginning and an end? $\endgroup$ – user160738 Jan 6 '16 at 10:52
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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

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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ '+' 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
  • 1
    $\begingroup$ $+$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.