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$\newcommand{\Ord}{\operatorname{Ord}}$When does it make sense to define the dimension of a space to be an infinite ordinal, instead of restricting to infinite cardinals?

We would essentially have to be paying attention to relations on the dimension set in addition to its cardinality, such as a well-ordering -- for example, we have that $\mathbb{R}^{\omega+1}$ has 'dimension' $\omega$ if we only pay attention to cardinality since $|\omega+1|=\omega$, but it can naturally be considered to have dimension $\omega+1$ since there is no order preserving bijection between $\omega+1$ and $\omega$.

While the trivial answer to the above appears to be 'if the dimension set is well-ordered', the discussion here leads me to believe that there are situations where the dimension set does not inherently come with a well-ordering, but can be well-ordered to canonically produce an ordinal dimension value.

Further confounding things is the fact that in the presence of choice any class can be well-ordered, which would indicate that we could always use an ordinal instead of a cardinal if the above trivial answer is correct. I suspect the correct answer is along the lines of 'if the dimension set has a definable well-ordering without choice', but this is just intuition. Any assistance or references would be appreciated.

EDIT: The nature of the space in question has intentionally been left vague; I am interested in any contexts or types of space for which an ordinal dimension makes sense.

SECOND EDIT: In light of the 'close as unclear' vote, here is a motivating example from a paper coming to the arxiv soon. Consider the ordinals $\Ord$ under natural (Hessenberg) operations as an ordered semialgebra over the natural numbers $\omega$ as an ordered semiring, and for all $\alpha\in O_n$ let $\alpha=\sum_{\gamma\in \Ord}\alpha_\gamma\omega^\gamma$ be the Cantor normal form of $\alpha$, so $|\{\gamma:\alpha_\gamma\neq0\}|<\omega$ and the exponents are increasing instead of decreasing. For all $\alpha\in \Ord$ define $$\downarrow_\alpha=\min\{\gamma:\alpha_\gamma\neq0\},$$ $$\downarrow_0=+\infty,$$ so $\downarrow_0>\downarrow_\alpha$ for all $\alpha\in \Ord\setminus\{0\}$ by definition, and then define an ordered subsemialgebra $\Ord_{\geq\omega^\beta}\subset \Ord$ at each $\gamma$-number $\omega^\beta$ by $$Ord_{\geq\omega^\beta}=\{\alpha\in \Ord:\downarrow_\alpha\geq\omega^\beta\}.$$ Then the ordered quotient space $\Ord_{\geq\omega^\alpha}/\Ord_{\geq\omega^{\alpha+1}}$ naturally admits a basis $$\{\omega^{\omega^\alpha+\sum_{i<\alpha}n_i\omega^i}+\Ord_{\geq\omega^{\alpha+1}}\}_{\{n_i\}_{i<\alpha}\in{^\alpha}\omega_{fin}},$$ where ${^\alpha}\omega_{fin}$ denotes the set of function from $\alpha$ into $\omega$ with finite support. This set is naturally well-ordered lexicographically with order type $\omega^\alpha$, so it makes sense to say that these semialgebras have ordinal dimension $\omega^\alpha$. As semimodules however, ignoring their inherited multiplicative structure, the canonical basis becomes $$\{\omega^{\sum_{i<\alpha+1}n_i\omega^i}+\Ord_{\geq\omega^{\alpha+1}}\}_{\{n_i\}_{i<\alpha+1}\in{^{\alpha+1}}\omega_{fin}},$$ where $n_\alpha>0$. This now has order type $\omega^{\alpha+1}$ in its lexicographic ordering and consequently a larger ordinal dimension. We can get the first coefficient using multiplication when viewing $\Ord_{\geq\omega^\alpha}/\Ord_{\geq\omega^{\alpha+1}}$ as a semialgebra, but we need to include those choices in the basis if we only have addition. Including exponentiation allows us to reduce the ordinal dimension of the algebra even further.

This example is obviously soaking in well-ordered structures so an ordinal dimension naturally makes sense, but hopefully it clarifies the nature of my question.

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  • $\begingroup$ Crossposted from MSE: math.stackexchange.com/questions/2757957/…. It received little attention for days, and the reference given in the comments is interesting but I would like a definition for spaces that do not inherently come with a topology. $\endgroup$ – Alec Rhea Apr 30 '18 at 22:32
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    $\begingroup$ It would help make your question less obscure if you clarified what you mean by "a space": a topological space? the spectrum of a commutative ring? something else? Or are you deliberately leaving the word "space" vague because you are interested in any kind of answer and it is, so to speak, part of the question? (The latter is legitimate, but, if so, you should still make it clear that you are deliberately making it vague. 😉) (contd.) $\endgroup$ – Gro-Tsen Apr 30 '18 at 22:37
  • $\begingroup$ (contd.) For all definitions of "space" I can think of right now, $\mathbb{R}^X$ depends, up to isomorphism, only on the cardinality of $X$, so that $\mathbb{R}^\omega$ and $\mathbb{R}^{\omega+1}$ are isomorphic (e.g., homeomorphic as topological spaces), and it makes no sense to try to ascribe them different invariants. $\endgroup$ – Gro-Tsen Apr 30 '18 at 22:39
  • $\begingroup$ @Gro-Tsen The vagueness was intentional, I'll edit to make that clear (no pun intended). How about as ordered vector spaces? I'll have to think some more on the topological homeomorphism, that is a good point. $\endgroup$ – Alec Rhea Apr 30 '18 at 22:42
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    $\begingroup$ Gro-Tsen's remark that $\mathbb R^X$ depends, up to isomorphism, only on the cardinality of $X$ is correct for all three of the product topology, the box topology, and the uniform topology. None of these refer to any order on $X$. $\endgroup$ – Andreas Blass Apr 30 '18 at 23:49
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A classic text is Dimension Theory by Hurevicz & Wallman (1941)

Among the types of dimension for topological spaces are the "upper inductive dimension" and the "lower inductive dimension". The form of the definition is such that they could be adapted to something with ordinal values. As I recall (it has been many years) Hurewicz & Wallman discuss the literature on this possibility in the notes at the end of one of their chapters.

Like this... For a topological space $X$ we say $\text{ind}(X) \le k$ iff there is a base $\mathcal B$ for the topology such that for all $V \in \mathcal B$, $\text{ind}(\partial V) \le k-1$. With $\text{ind}( \varnothing) = -1$ this is conventionally used to define the small inductive dimension for natural numbers $k$. $\text{ind}(X) = k$ iff $k$ is the least natural number such that $\text{ind}(X) \le k$.

But it will also work for transfinite ordinals. Because of the inductive form of the definition.

$\text{ind}(X) \le \omega$ iff there is a base $\mathcal B$ for the topology such that for all $V \in \mathcal B$, $\text{ind}(\partial V) < \omega$. $\text{ind}(X) \le \omega+1$ iff there is a base $\mathcal B$ for the topology such that for all $V \in \mathcal B$, $\text{ind}(\partial V) \le \omega$. And in general, $\text{ind}(X) \le \gamma$ iff there is a base $\mathcal B$ for the topology such that for all $V \in \mathcal B$, $\text{ind}(\partial V) < \gamma$.

Similar discussion for the large inductive dimension.

How interesting this turns out to be, I do not recall.

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  • $\begingroup$ Thank you for the reference! I recently ordered a copy of Theory of Dimensions: Finite and Infinite by Ryszard Engelking which I believe is a more recent treatise on these same subjects, but this book will make a nice complement at amazons price point :). $\endgroup$ – Alec Rhea May 1 '18 at 18:22
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One natural context where one thinks of the exponent or dimension in the space as having an order is with the function space $\kappa^\lambda$ considered under the relation of eventual domination. That is, for functions $f,g:\lambda\to\kappa$, we say $f\leq^*g$ just in case there is some $\beta<\lambda$ such that for all $\alpha\in[\beta,\lambda)$ we have $f(\alpha)\leq g(\alpha)$. Here, it is the order of $\lambda$ that matters and not just its cardinality.

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  • $\begingroup$ Very cool -- this seems a situation similar to my example above where the dimension set is literally a well-ordered set, so an ordinal dimension makes sense. I wonder if this can be generalized to dimension sets with a partial order on them, perhaps as the largest ordinal that embeds in the partial order or the length of the largest antichain? $\endgroup$ – Alec Rhea May 1 '18 at 18:25

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