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I worked this theory : A new generalization of the dimension?

I have a theorem about dimensions which is more general and simple than for matroids.

Definition 1: A structure $S$, is a pair $(X, \mathcal T)$ with $X$ a set and $\mathcal T$ a set of subsets of $X$ which is stable with respect to arbitrary intersections, with $X \in \mathcal T$.

Definition 2: For $U \subset X$, we denote $\langle U\rangle_S:=\bigcap \limits_{F \in \mathcal T, U \subset F} F$.

Definition 3: For a structure $S=(X,\mathcal T)$, we say the set $U\neq \emptyset$ is free if $$ \forall u \in U,\ u \notin \langle v \mid v \in U,v\neq u \rangle_S $$

Definition 4: For a structure $S=(X,\mathcal T)$, we say this structure has a dimension if $\forall U \subset X$ free and $v \notin \langle U\rangle_S$, the set $U \cup \{v\}$ is free.

Definition 5: For a structure $S=(X,\mathcal T)$ and $F \in \mathcal T$, we denote $\dim(F)=n$ if the largest free set of $F$ has a cardinality of $n$.

Theorem 1: For a structure $S=(X,\mathcal T)$ with a dimension and $E,F \in \mathcal T$, if $\dim(E)=\dim(F)<\infty$ and $E \subset F$ then $E=F$

Theorem 2: For a structure $S=(X,\mathcal T)$ with $V \subset\langle U\rangle_S$, if $\text{card}(U)<\text{card}(V)$ then $V$ is not free.

Example 1: $S=(\mathbb R,\mathcal F)$, the closed sets of reals, is a structure with a dimension, and the $\dim(\mathbb R)=\text{card}(\mathbb N)$, because if $A \subset \mathbb R$ with $\text{card}(A)>\text{card}(\mathbb N)$ then it exists $(a_n) \in A^{\mathbb N}$ injective with $\lim a_n=c$ and $c \in A$.

Example 2: $S=(X=C([0,1],\mathbb R),\mathcal F)$, the closed sets for the uniform norm $||.||_{\infty}$, it's a structure with a dimension, we know $\langle \mathbb Q[x] \rangle_S=X$ and $\text{card}(\mathbb Q[x])=\text{card}(\mathbb N)$, so by theorem2, if $V \subset X$ with $\text{card}(V)>\text{card}(\mathbb N)$ then $V$ has an accumulation point, for the uniform norm.

Question: Is this generalization of the dimension already known?

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    $\begingroup$ (Also, it seems that the actually important object here is the closure operator $cl_S: U\mapsto\langle U\rangle_S$; just talking about the pair $(X, cl_S)$ seems much easier than talking about $(X,\mathcal{T})$, and also makes the relationship with matroids etc. clearer.) $\endgroup$ – Noah Schweber Jan 14 at 0:54
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    $\begingroup$ It seems that a matroid gives rise to a structure with a dimension, by taking $\mathcal T$ to consist of the closed sets (the flats) of the matroid. In particular, your notion of "having a dimension" seems to correspond to the exchange property of matroids. Is that correct? Do you have examples of structures with dimension that don't come from matroids in this way? $\endgroup$ – Andreas Blass Jan 14 at 1:38
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    $\begingroup$ Note that what you call a "structure" is also known as a "Moore family": en.wikipedia.org/wiki/Closure_operator $\endgroup$ – j.c. Jan 14 at 1:43
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    $\begingroup$ @Vincent E. g. the set of subspaces of a vector space. $\endgroup$ – მამუკა ჯიბლაძე Jan 14 at 10:05
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    $\begingroup$ Responding to your edit: Your Theorem 2 is not correct, I've added a counterexample at the bottom of my answer. $\endgroup$ – Alex Kruckman Jan 21 at 21:30
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Let me try to situate your definition in context.

A closure operator on a set $X$ is a function $\text{cl}\colon \mathcal{P}(X)\to \mathcal{P}(X)$ such that for all $A,B\subseteq X$, we have:

  1. $A\subseteq \text{cl}(A)$.
  2. If $A\subseteq B$, then $\text{cl}(A) \subseteq \text{cl}(B)$
  3. $\text{cl}(\text{cl}(A)) = \text{cl}(A)$.

We say the closure operator satisfies exchange if additionally for all $A\subseteq X$ and $a,b\in X$:

  1. If $b\in \text{cl}(A\cup \{a\})$ but $b\notin \text{cl}(A)$, then $a\in \text{cl}(A\cup \{b\})$.

When $X$ is finite, a pair $(X,\text{cl})$ such that $\text{cl}$ is a closure operator on $X$ satisfying exchange is called a (finite) matroid [this is one of many equivalent definitions of this concept].

Now we can clearly make the same definition when $X$ is infinite, but it turns out not to have some of the nice properties of matroids. For example, we can no longer prove that every closed set has a basis (a maximal independent subset). So usually people strengthen this definition by adding extra axioms: it's common to look at finitary matroids (also called a pregeometries, especially by model theorists) or B-matroids. See Wikipedia.


Your definition is a slight weakening of the notion of a closure operator satisfying exchange, which becomes equivalent to this notion if we add an assumption about existence of bases for closed sets.

First let's observe that what you call a structure is equivalent to a closure operator.

In one direction, given a structure $S = (X,\mathcal{T})$, we can define $\text{cl}(A) = \langle A\rangle_S$ as in your question, and this clearly satisfies axioms 1, 2, and 3.

Conversely, given a closure operator $\text{cl}$ on $X$, we can take $\mathcal{T}$ to be the set of closed subsets of $X$, i.e. $\mathcal{T} = \{A\subseteq X\mid \text{cl}(A) = A\}$. Then $\mathcal{T}$ is stable under arbitrary intersections, since if $A_i\in \mathcal{T}$ for all $i\in I$, then $\bigcap_{i\in I} A_i \subseteq A_i$ for all $i\in I$, so $$\text{cl}(\bigcap_{i\in I} A_i)\subseteq \text{cl}(A_i) = A_i$$ for all $i\in I$, and hence $$\text{cl}(\bigcap_{i\in I} A_i) \subseteq \bigcap_{i\in I} A_i.$$ In the other direction, we have $$\bigcap_{i\in I} A_i \subseteq \text{cl}(\bigcap_{i\in I} A_i),$$ so these sets are equal, and $\bigcap_{i\in I} A_i\in \mathcal{T}$.


Now if $\text{cl}$ satisfies exchange, then the structure $(X,\mathcal{T})$ has a dimension in your sense.

Assume $U\subseteq X$ is free and $v\notin \text{cl}(U)$. Assume for contradiction that $U\cup \{v\}$ is not free. Then since $v\notin \text{cl}(U)$, we must have some $u\in U$ such that $u\in \text{cl}(U'\cup \{v\})$, where $U' = U\setminus \{u\}$. But since $U$ is free, $u\notin \text{cl}(U')$. So by exchange, $v\in \text{cl}(U'\cup \{u\}) = \text{cl}(U)$, contradiction.


We can prove the converse, if we additionally assume existence of bases. Let's say that a structure $S = (X,\mathcal{T})$ has bases if for every set $A\subseteq X$, there is a free set $A'\subseteq X$ such that $\text{cl}(A') = \text{cl}(A)$.

Now suppose $S = (X,\mathcal{T})$ is a structure with a dimension and bases. To prove exchange, suppose $b\in \text{cl}(A\cup \{a\})$ but $b\notin \text{cl}(A)$.

Let $A'$ be a free set such that $\text{cl}(A') = \text{cl}(A)$. Then $b\notin \text{cl}(A')$, so since $S$ has a dimension, $A'\cup \{b\}$ is free. Now $(A'\cup \{b\})\cup \{a\}$ is not free (since $b\in \text{cl}(A'\cup \{a\}) = \text{cl}(A\cup \{a\})$), so since $S$ has a dimension, $a\in \text{cl}(A'\cup \{b\}) = \text{cl}(A\cup \{b\})$.


Here is a counterexample showing that not every structure with a dimension has basis, and the associated closure operator does not satisfy exchange in general.

Let $X = A \cup \{a,b\}$, where $A$ is infinite and $a,b\notin A$. Define a structure by letting $\mathcal{T} = \mathcal{P}_{\text{fin}}(X) \cup \{A\} \cup \{A\cup \{b\}\}\cup \{X\}$, where $\mathcal{P}_{\text{fin}}(X)$ is the set of all finite subsets of $X$. This family is evidently stable under arbitrary intersections.

The corresponding closure operator has $\text{cl}(B) = B$ if $B$ is finite, and $\text{cl}(B)$ is the least of $A$, $A\cup \{b\}$ and $X = A\cup \{a,b\}$ containing $B$ if $B$ is infinite.

A subset $B\subseteq X$ if free if and only if it is finite. Indeed, any infinite subset $B\subseteq X$ contains infinitely many elements of $A$, and for any $z\in B\cap A$, we have $z\in \text{cl}(B\setminus \{z\})$, since $A\subseteq \text{cl}(B\setminus \{z\})$.

So this structure has a dimension, since for any free $U$ and any $v\notin \text{cl}(U)$, $U\cup \{v\}$ is finite, hence free.

But the closed set $A$ has no basis (since any free set has finite closure). And exchange fails: we have $b\in \text{cl}(A\cup \{a\})$ and $b\notin \text{cl}(A)$, but $a\notin \text{cl}(A\cup \{b\}$.


Your definition of "has a dimension" is enough to get bases for closures of finite sets $A$ (just build up a free set by adding elements of $A$ not in the closure of what we've built so far one by one, until the closure of our basis contains $A$ and hence is equal to the closure of $A$). So it's enough to get exchange when the set $A$ appearing in the definition of exchange is finite. The problem arises when you try to "take limits": the union of an increasing sequence of free sets is not necessarily free.

This is one motivation for the definition of finitary matroid. If we assume that $\text{cl}$ is finitary (this is equivalent to assuming that the family $\mathcal{T}$ is stable under directed unions), then "has a dimension" will imply the existence of bases and exchange.

That is, a structure $(X,\mathcal{T})$ with a dimension such that $\mathcal{T}$ is stable under directed unions, is essentially the same thing as a finitary matroid.


Conclusion: I doubt that your exact definition has been studied before. Usually people are interested in strengthenings of the notion of closure operator satisfying exchange. Your notion is a weakening, and as the counterexample above shows, without some extra hypotheses, the "has a dimension" assumption can trivialize when looking at closures of infinite sets, simply because not enough free sets exist.

By the last remark above, the finite dimensional sets in your structures with dimension will behave just like they do in finitary matroids. Thus you'll be able to prove things like the theorem in your question. But without any additional assumptions on how closures of infinite sets behave, I suspect you'll have a hard time proving anything interesting about the infinite dimensional sets.


Edit: Your Theorem 2 is not correct.

For a counterexample, let $X = \mathbb{R}$, and let $\mathcal{T}$ be the set of all $A\subseteq \mathbb{R}$ such that $A = \mathbb{R}$ or $A\cap \mathbb{N}$ is finite. This family is stable under arbitrary intersections, and the structure $(\mathbb{R}, \mathcal{T})$ has a dimension, since $U\subseteq \mathbb{R}$ is free if and only if $U\cap \mathbb{N}$ is finite, so the union of any free set with any singleton is free.

Now let $V = \mathbb{R}\setminus \mathbb{N}$. We have $\langle \mathbb{N}\rangle = \mathbb{R}$, so $V\subseteq \langle \mathbb{N}\rangle$, and $\text{card}(\mathbb{N})< \text{card}(V)$, but $V$ is free, since $V\cap \mathbb{N} = \emptyset$.

If you were to explain your supposed proof of Theorem 2, I might be able to find where the proof goes wrong and suggest a strenthening of your axioms that makes it true.

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