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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.

$\newcommand{\Ord}{\operatorname{Ord}}$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.

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 $$\{\omega^{\omega^\alpha+\sum_{i<\alpha}n_i\omega^i}\}_{\{n_i\}_{i<\alpha}\in{^\alpha}\omega_{fin}},$$ 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 $$\{\omega^{\sum_{i<\alpha+1}n_i\omega^i}\}_{\{n_i\}_{i<\alpha+1}\in{^{\alpha+1}}\omega_{fin}},$$ 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.

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.

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.

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 $$\{\omega^{\omega^\alpha+\sum_{i<\alpha}n_i\omega^i}\}_{\{n_i\}_{i<\alpha}\in{^\alpha}\omega_{fin}},$$ 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 $$\{\omega^{\sum_{i<\alpha+1}n_i\omega^i}\}_{\{n_i\}_{i<\alpha+1}\in{^{\alpha+1}}\omega_{fin}},$$ 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.