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For a given set $X$, what is the notation for the space of all finite $X$-valued sequences? I realise that the space of $n$-tuples is written as $X^n$, and the space of infinite sequences is $X^\mathbb{N}$, i.e. the space of functions from $\mathbb{N}$ to $X$. But how do you denote the space of arbitrary, finite sequences in $X$?

I'm thinking it should be some kind of direct limit, but how?

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3 Answers 3

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In set theory, $\omega$ is the ordinal which represents $\Bbb N$, and it is common to write $X^{<\omega}$ as the set of all finite sequences. I suppose that one can write $X^{<\Bbb N}$ as well.

I have also seen $\operatorname{Seq}(X)$ being used for this purpose. And you can always introduce a notation which seems as if it makes sense (e.g. one of the above, if they don't seem like a standard notation to you).

Let $X$ be a set, we denote by ... the set of all finite sequences of elements from $X$.

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I like $X^{<\omega}$. I hadn't noticed it so far. It seems to be pretty self-explanatory. –  Kerrek SB Jun 13 at 23:26
    
It's quite common in set theory, and more generally $X^{<\kappa}$ where $\kappa$ is a cardinal. –  Asaf Karagila Jun 13 at 23:26

I think I see it now. It's not concise notation or anything I've seen in widespread use, but at least a formal definition should look something like this:

Let $Y_n = \coprod_{i=1}^n M^{\times i}$. Then there is a directed system $Y_i \to Y_{i+1}$. The space of finite sequences in $M$ should be the direct limit of this system, $\varinjlim_{n} Y_n$.

Update: Actually, $\varinjlim_{n \to \infty} Y_n$ is simply $\coprod_{n=0}^\infty M^{\times n} = \varnothing \sqcup M \sqcup M^2 \sqcup \dotsb$

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Since you used the category theory tag, I'll add that finite sequences make sense in a more general setting than sets, namely in a topos with a natural numbers object. In that case the object of finite sequences from $X$ is called the $\textit{list object}$ on $X$. Different authors use different notation for this: Johnstone uses $L(X)$, Vickers uses $List(X)$, and I've even seen $X^{[n]}$.

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The last one looks frightfully like a Hilbert scheme of points, though :-S –  Kerrek SB Jun 14 at 0:53

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