This theorem seems "obvious" to me, but I want to check my logic since I am un-familiar with un-countably infinite sets, and I know these can give rise to non-intuitive results. Any comments welcome.


The cardinality of a set $A$ will be denoted $|A|$. If $|A|=|\mathbb{N}|$, $A$ will be called countable or countably infinite and this will be denoted $|A|=\aleph_{0}$. If $A$ is infinite and $|A|\ne\aleph_{0}$, $A$ will be called uncountable or uncountably infinite.


If $A$ is un-countable then for any $a\in A$, $A$ contains an uncountable number of disjoint co-countable sets that contain $a$. More formally;

For any $a\in A$ where $|A|\not=\aleph_{0}$ there exist subsets $B_{\beta}\subset A$, $\beta\in I_{\beta}$ for an index set $I_{\beta}$, $|I_{\beta}|\not=\aleph_{0}$, such that for all $\beta\in I_{\beta}$ : $a\in B_{\beta}$ and $|A-B_{\beta}|=\aleph_{0}$. Furthermore for any $\beta\not=\beta'$, $B_{\beta}\cap B_{\beta'}=\emptyset$


Without loss of generality let $I_{\beta}=\mathbb{R}$ and choose $\beta\in I_{\beta}$. Choose $a_{\beta,1}\in A$, $a_{\beta,1}\not=a$. Since $|A-\{a_{\beta,1}\}|\not=\aleph_{0}$ there exists $a_{\beta,2}\in\{A-\{a_{\beta,1}\}\}$, $a_{\beta,2}\not=a$. Since $|A-\{a_{\beta,1},a_{\beta,2}\}|\not=\aleph_{0}$ there exists $a_{\beta,3}\in\{A-\{a_{\beta,1},a_{\beta,2}\}\}$, $a_{\beta,3}\not=a$. Continuing in this way $n_{\beta}$ times, $n_{\beta}\in\mathbb{N}$, we can construct a set $B_{\beta}=A-\{a_{\beta,1},a_{\beta,2},...,a_{\beta,n_{\beta}}\}$ that satisfies $a\in B_{\beta}\subset A$ and $|A-B_{\beta}|=|\{a_{\beta,1},a_{\beta,2},...,a_{\beta,n_{\beta}}\}|=\aleph_{0}$ so that $B_{\beta}$ is a co-countable subset of $A$ containing $a$.

Choose another $\beta'\in I_{\beta}$, $\beta\not=\beta'$. Since $|B_{\beta}|\not=\aleph_{0}$ there exists $a_{\beta',1}\in B_{\beta}$, $a_{\beta',1}\not=a$. Since $|B_{\beta}-\{a_{\beta',1}\}|\not=\aleph_{0}$ there exists $a_{\beta',2}\in B_{\beta}-\{a_{\beta',1}\}$, $a_{\beta',2}\not=a$. Since $|B_{\beta}-\{a_{\beta',1},a_{\beta',2}\}|\not=\aleph_{0}$ there exists $a_{\beta',3}\in B_{\beta}-\{a_{\beta',1},a_{\beta',2}\}$, $a_{\beta',3}\not=a$. Continuing in this way $n_{\beta'}$ times, $n_{\beta'}\in\mathbb{N}$, we can construct a set $B_{\beta'}=B_{\beta}-\{a_{\beta',1},a_{\beta',2},...,a_{\beta',n_{\beta'}}\}$ that satisfies $a\in B_{\beta'}\subset A$ and $|A-B_{\beta'}|=|\{a_{\beta,1},a_{\beta,2},...,a_{\beta,n_{\beta}}\}\cup\{a_{\beta,1},a_{\beta,2},...,a_{\beta',n_{\beta'}}\}|=\aleph_{0}+\aleph_{0}=\aleph_{0}$ so that $B_{\beta'}$ is a co-countable subset of $A$ containing $a$.

Since $B_{\beta}\cap B_{\beta'}=\emptyset$ and since we can choose un-countably many $\beta\in I_{\beta}$ the result is proved $\hspace{5pt}\blacksquare$

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    $\begingroup$ Your definition of "uncountably infinite" is not standard. Usually, it is an infinite set $S$ such that $|S|\neq \aleph_0.$ There are a lot of uncountably infinite cardinals other than $|\mathbb R.|$ $\endgroup$ – Thomas Andrews Mar 27 at 17:31
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    $\begingroup$ No, you are wrong, @AndrésE.Caicedo Given $a\in A$ take a countable subset $C\subseteq A$ with $a\not\in C.$ Then take $\{A\setminus T\mid T\subset C, \text{ and T infinite}\}$. There are uncountably many (countably) infinite subsets $T$ of $C$, and for each $a\in A-T,$ and $A-T$ is co-countable. $\endgroup$ – Thomas Andrews Mar 27 at 18:03
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    $\begingroup$ @Thomas These sets are not disjoint. $\endgroup$ – Andrés E. Caicedo Mar 27 at 18:05
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    $\begingroup$ @dandar Without the requirement of disjointness, of course the claim holds, and it is straightforward. Thomas sketched in comments what to do. I do not see uncountably many sets produced using procedures along the lines your comment suggests. If besides disjointness we also ignore the requirement of having uncountably many such sets, then yes, your comment gives a possible approach. $\endgroup$ – Andrés E. Caicedo Mar 28 at 16:27
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    $\begingroup$ @dandar Yes, that's the argument. $\endgroup$ – Andrés E. Caicedo Mar 28 at 18:33

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