Let $\mathcal{A}$ be an uncountable collection of sets so that if $I_{\mathcal{A}}$ is the index set of elements of $\mathcal{A}$ then $|I_{\mathcal{A}}|\not=\aleph_{0}$ (I mean this to mean not finite as well so perhaps I should use $|I_{\mathcal{A}}|>\aleph_{0}$?). The underlying set is $\Omega$, and each $A\in\mathcal{A}$ satisfies $\omega\in A\subset\Omega$ and $|\Omega-A|=\aleph_{0}$ so that $\mathcal{A}$ is an uncountable collection of co-countable sets containing the point $\omega$.

Can any form of indexing be used to fully enumerate all sets in $\mathcal{A}?$

Since $\{A_{i}\}_{i\in \mathbb{N}}\subset \mathcal{A}$ then obviously I cannot use a countable index set to fully enumerate the sets in $I_{\mathcal{A}}$. The trivial index set $I_{\mathcal{A}}=\mathcal{A}$ permits use of any symbol, say $i$, so that $i\in\mathcal{A}\Longleftrightarrow A_{i}\in \mathcal{A}$, and this might need to be formalised with a bijection $f:I_{\mathcal{A}}\longrightarrow \mathcal{A}$ such that $f(i)\longmapsto A_{i}$. If this type of indexing makes sense then I am tempted to write $\mathcal{A}:=\{A_{\alpha}\}_{\alpha\in I_{\mathcal{A}}}$ where I use Greek letters like $\alpha$ instead of $i,j,k$ etc., but of course this is just my preference.

Does $\bigcap_{\alpha\in I_{\mathcal{A}}}A_{\alpha}=\{\omega\}$ hold without imposing further conditions on the $\mathcal{A}$-sets?

The result would obviously hold if $\{\omega\}\in\mathcal{A}$. However $|\Omega-\{\omega\}|\not=\aleph_{0}$ implies $\{\omega\}\not\in\mathcal{A}$. My intuition then is no the claim is false without further conditions. I am particularly interested in the condition $\alpha<\alpha+1$ for all $\alpha\in I_{\mathcal{A}}$ if and only if $A_{\alpha}\supset A_{\alpha+1}$, so that $\mathcal{A}$ is a class of decreasing co-countable sets (see next section for objections to this ordering notation).

My proof: By definition $\omega\in A_{\alpha}$ for all $A_{\alpha}\in\mathcal{A}$ so that $\{\omega\}\subseteq\bigcap_{\alpha\in I_{\mathcal{A}}}A_{\alpha}$. Choose $\omega'\in \bigcap_{\alpha\in I_{\mathcal{A}}}A_{\alpha}$ such that $\omega'\not=\omega$. By the decreasing nature of the $\mathcal{A}$-sets there exists a $\alpha'\in I_{\mathcal{A}}$ such that $\omega'\not\in A_{\beta}$ for all $\beta>\alpha'$. This contradicts $\omega'$ being contained in the intersection of all sets. Thus $\omega'=\omega$ which implies $\bigcap_{\alpha\in I_{\mathcal{A}}}A_{\alpha}\subseteq\{\omega\}$ and the result follows.

I understand there are various "intersection theorems" that give $\bigcap_{\alpha\in I_{\mathcal{A}}}A_{\alpha}=\{\omega'\}$ for some $\omega'\in\Omega$ which implies $\omega'=\omega$ in my set-up here. These theorems use conditions such as the $\mathcal{A}$-sets being non-empty, closed and compact, or alternatively that $\Omega$ is complete where the $\mathcal{A}$-sets are non-empty, bounded, closed, and whose diameters converge to zero.

Do I need to define a well-ordering of $\mathcal{A}$ and $I_{\mathcal{A}}$ and use order types?

At least one possible problem with the above proof (other than possibly the logic) is the concept of ordering of both $\mathcal{A}$ and $I_{\mathcal{A}}$, which I rather lazily assume can be always achieved (Zermelo's Well-Ordering Theorem lets me off the hook?), but which I am struggling to determine if I need to justify. By an ordering I mean the following: $\alpha+1$ is the next index to $\alpha$ indexing $A_{\alpha+1}$ which is the next set to $A_{\alpha}$. Conversely $A_{\alpha+1}$ is the next set to $A_{\alpha}$ implies $(\alpha+1)$ is the next index to $\alpha$. Thus $\alpha<\alpha+1\Longleftrightarrow A_{\alpha}<A_{\alpha+1}$. The notation $A_{\alpha}<A_{\alpha+1}$ is to be defined but I have in mind $A_{\alpha}\subset A_{\alpha+1}$. By a well-ordering of I mean that every non-empty subset of $I_{\mathcal{A}}$ and $\mathcal{A}$ have a first element. If this makes sense for an uncountable set or class of sets, then the countable indices I have used feels wrong to me since again $\{A_{\alpha_{n}}\}_{n\in\mathbb{N}}\subset\mathcal{A}$.

For this reason I am wondering if my ordering notation needs to be formalised with order types and comparison of ordinal numbers? In this respect, for an uncountable set $A=\{a_{0},a_{1},...,a_{\omega},a_{\omega+1},....\}$ where $\omega$ is the order type of $\mathbb{N}$ with the natural ordering ($\omega$ here is not the same as $\omega\in\Omega$ in previous sections), $a_{2}$ has index "2" since the subset $\{a_{0},a_{1}\}$ has order type $2$ which means it can be brought into a 1-1 correspondence with the set $\{1,2\}$ which is defined to have order type 2. Similarly $a_{\omega}$ has index $\omega$ since the subset $\{a_{0},a_{1},...\}$ has order type $\omega$, meaning it can be brought into a 1-1 correspondence with $\mathbb{N}$. The element $a_{\omega+1}$ has index $\omega+1$ since the subset $\{a_{0},a_{1},...,a_{\omega}\}$ can be brought into a 1-1 correspondence with sets of the type $\{0,1,...,a\}$ which is defined to have order type $\omega+1$.

Using the above order types, $0<1$ means $\{a_{0}\}$ is similar to a subset of $\{a_{0},a_{1}\}$, i.e. $\{a_{0}\}\approxeq\{a_{0}\}$ in the sense a 1-1 mapping exists (the identity function) preserving the ordering of the elements. In this way $(n-1)<n$ since $\{a_{0},a_{1},...,a_{n-1}\}\approxeq \{a_{0},a_{1},...,a_{n-1}\}\subset\{a_{0},a_{1},...,a_{n-1},a_{n}\}$. Similarly $n<\omega$ since $\{a_{0},a_{1},...,a_{n}\}\approxeq \{a_{0},a_{1},...,a_{n}\}\subset\{a_{0},a_{1},...,a_{n-1},a_{n},...\}$. These subsets are called sections of $A$, denoted $S_{A}(a)$ determined by an $a\in A$: $S_{A}(a):=\{a'\in A:a'<a\}$, i.e. $\{a_{0},a_{1},...,a_{n-1}\}=S_{A}(a_{n})$. This gives rise to the definition of $\alpha<\alpha+1$ for a general order type $\alpha$ as meaning $A_{\alpha}$ is similar to one of the sections of $A$.

I presume the above considerations can be applied to well-ordered uncountable classes as well as sets?


To index an uncountable set use ordinal numbers, preferablely initial ordinals.

Often the use of indexing makes for needless effort.
Let C be an uncountable collection of sets.
The union of those sets is
$\cup$C = { x : exists A in C with x in A }.

I leave for the reader the chore of indexing C and defining the union using indexing. Which is preferable?

If S is an uncountable set and C a collection of cocountable subsets of S and
for all A in C, x in A, then x in $\cap$C.
If C is the collection of all the cocountable subsets of S and
for all A in C, x in A, then {x} = $\cap$C.

Indexing is not needed for that problem. If indexing is imperative, well ordeing of the index isn't needed, only that the index set is equinumerous to the set being index. As demonstrated above, the collection of sets itself can be used for the indexing set which renders the indexing moot.

Indexing is needed, among other things, for large monotonic sets.
Let C be an uncountable collection of sets and I be the initial ordinal with cardinality C. C = { C$_j$ : j in I } is a strictly increasing collection of sets when for all j,k in I, (j < k implies $C_j \subset C_k$).

  • $\begingroup$ Thank you very much for this clear answer. I am very happy to avoid indexing! The confirmation of the proof result not needing such indexing is the key for me. Thanks too for the guidance on initial ordinals. $\endgroup$ – dandar Apr 12 at 7:37

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