$\newcommand{\I}{\mathcal I}$$\newcommand{\N}{\mathbb N}$$\newcommand{\Q}{\mathbb Q}$$\newcommand{\R}{\mathbb R}$In this post I’m going to explore partitioning the unit interval $\I := [0,1]$ into uncountably many pieces where each piece is also uncountable.
Countably many uncountable pieces
Just to get started, it is not hard to partition $\I$ into countably many uncountable sets. One such partition is given by
$$
S_1 = \{0\}\cup\left(\frac 12, 1\right]
$$
and
$$
S_n = \left(\frac{1}{n+1}, \frac{1}{n}\right]
$$
for $n = 2,3,\dots$. This collection is indexed by $\N$ so it’s countable. Letting $\lambda$ be the Lebesgue measure, for each $n\in\N$
$$
\lambda(S_n) = \frac 1n – \frac 1{n+1} = \frac{1}{n(n+1)}
$$
which is strictly positive. If $A$ is a countable subset of $\mathbb R$ then
$$
\lambda(A) = \sum_{a\in A}\lambda(\{a\}) = 0
$$
by $\sigma$-additivity, i.e. being countable implies having measure zero. The contrapositive of this is that having positive measure means the set in question is uncountable, therefore $\{S_n : n\in\N\}$ gives a partition of $\I$ into countably many pieces where each one has uncountably many elements.
Uncountably many countable pieces
Next I’m going to partition $\I$ into uncountably many pieces each of which has countably many elements.
For this I will think of the group $(\R, +)$ and then $(\Q, +)$ is a normal subgroup. I can consider the quotient group $\R / \Q$ which consists of cosets of the form $x + \Q$ for $x \in \R$. For a given $x\in \R$ I will use $[x]$ to denote its coset.
First, note that since any coset is a translation of $\Q$ it is dense in all of $\R$. This also means each coset is countable, and since $\bigcup_{H \in \R/ \Q} H = \R$ this means that there are uncountably many cosets in $\R / \Q$.
The actual partition that I’ll consider is $\{H \cap \I : H \in \R / \Q\}$. Since each coset $H$ is dense (but also countable) in $\I$ I know that there are still countably many elements to $H \cap \I$ so this restriction to $\I$ isn’t a problem.
Digression: Vitali sets
I’m going to digress and discuss Vitali sets here since that’s the first place that I came across this particular partition of $\I$; I’ll be following the proof given in the wikipedia article. If I invoke the Axiom of Choice ($\mathsf{AC}$) I can create a set $V$ where $V$ contains exactly one element of each coset. I claim $V$ is not Lebesgue measurable. Note $V$ is uncountable as it’s one-to-one with $\R/\Q$ and for any $u,v \in V$ it must be that $u-v$ is irrational else $u$ and $v$ would be in the same coset.
Pf of non-measurability: Suppose it is measurable. Let $q_1,\dots$ enumerate $\Q\cap[-1,1]$.
For each $k\in\N$ consider $V_k := q_k + V$, i.e. shift $V$ by $q_k$. I’ll now show that the $V_k$ are pairwise disjoint. Suppose there is an $x \in V_i\cap V_j$ for $i\neq j$. Then $x = v + q_i = w + q_j$ where $v, w \in V$. $q_i \neq q_j$ by construction and this also means $v \neq w$. Then
$$
(v – w) + (q_i – q_j) = 0 \implies v – w = q_j – q_i\in \Q
$$
but $v, w \in V$ means their difference is irrational so this is a contradiction, and therefore $V_i\cap V_j = \emptyset$.
Next I’ll show that $[0,1] \subseteq \bigcup_{k \in \N} V_k \subseteq [-1,2]$. The RHS is easy as $V \subset [0,1]$ and I’m shifting $V$ by a rational in $[-1,1]$ for each $V_k$. For the LHS, let $r \in [0,1]$. My cosets partition $[0,1]$ so $r$ belongs to some equivalence class $[r]$. Let $v$ be the representative of this equivalence class in $V$. Then $r, v \in [r]$ so $r -v = q_i$ for some $i\in\N$; in other words, $r = v + q_i \in V_i$. Thus $\I \subseteq \bigcup_{k \in \N} V_k \subseteq [-1,2]$.
To finish the proof, by $\sigma$-additivity I have
$$
\lambda([0,1]) = 1 \leq \sum_{k\in\N} \lambda(V_k) \leq 3
$$
but the $V_k$ are all translations of $V$ so $\lambda$ assigns them the same measure. This means
$$
1 \leq \sum_{k\in\N} \lambda(V) \leq 3
$$
but this is impossible as the infinite sum of a non-negative constant is either $0$ or $\infty$. Thus $\lambda(V)$ can’t exist.
$\square$
This result gives a sense of just how fractured $\R/\Q$ is. But can I go further?
Uncountably many uncountable sets
The answer is yes. I can do a clever reindexing and combine some cosets to turn $\R/\Q$ into an uncountable union of uncountable sets. I learned of this from Brian M. Scott’s answer on this Math.SE post.
With $\R / \Q$ I have $\mathfrak c$-many cosets (where $\mathfrak c = |\R|$). In terms of cardinality $\vert\R\vert = \vert\R^2\vert$ so I could index these cosets with pairs of real numbers. Thus I have an uncountable collection $\{H_{(x,y)} : (x,y)\in \R^2\}$ of countable disjoint sets. I can now let
$$
S_x = \bigcup_{y \in \R} H_{(x, y)}
$$
which is an uncountable union of disjoint countable sets so each set $S_x$ is uncountable, but I also still have uncountably many of them. This means $\{S_x : x\in\R\}$ is a partition of $\R$ into uncountably many uncountable sets, so I just need to restrict this to $\I$ and I’m good to go.
Consider a bijection $\varphi : \R \to \I$ and then use this to map these $S_x$ to a partition of $\I$. This means I’m considering
$$
\mathcal P = \{\varphi(S_x) : x \in \R\}
$$
as my partition. Does this do the trick?
Suppose $z \in \varphi(S_x)\cap\varphi(S_y)$ for $x\neq y$. Then there is some $u \in \varphi(S_x), v\in \varphi(S_y)$ such that $z = u = \varphi(s_x)$ and $z = v = \varphi(s_y)$. But this means $\varphi(s_x) = \varphi(s_y) \implies s_x = s_y$ but that contradicts the $\{S_x\}$ being a disjoint. So it must be that the $\{\varphi(S_x)\}$ are still disjoint.
Now do they cover $\I$? Let $r \in \I$. $\varphi$ is a bijection from $\R$ to $\I$ so $\varphi^{-1}(r)$ is some element of $\R$, which means there is a $y \in \R : \varphi^{-1}(r) \in S_y$. Applying $\varphi$ means $r \in \varphi(S_y)$.
Thus $\{\varphi(S_x) : x \in \R\}$ is indeed a partition of $\I$ into uncountably many uncountable pieces.
$\square$
A different unfinished approach
To conclude, I’m going to write up a bit of one approach that I explored but didn’t finish (as I found Brian Scott’s solution and chose to use that instead). I got this idea from here.
I managed to get an uncountable collection of cosets by taking the quotient of $(\R, +)$ w.r.t. a countable normal subgroup $(\Q, +)$ but a consequence was that each coset was countable. This suggests that the next thing to do is to instead consider $\R / G$ where $G$ is an uncountable subgroup of $\R$. But how to even get such a group? I’m freely using $\mathsf{AC}$ so I’ll use that to produce such a group $G$ although this will make $G$ very non-constructive.
Consider $\R$ as a vector space over $\Q$. Assuming $\mathsf{AC}$ it must have a Hamel basis $B = \{\xi\}$ so any real number $r \in \R$ can be written as
$$
r = q_1\xi_1 + \dots + q_n\xi_n
$$
where the $q_j \in \Q$. Furthermore this representation is unique (by definition of a basis). It’s also the case that $B$ is uncountable.
I can see this by supposing $B$ to be countable. By construction $\R$ is the $\Q$-span of $B$. I can think of $\R$ being generated as follows. First, I’ll consider $F := \mathcal P_{\text{fin}}(B)$, i.e. $F$ is the finite powerset of $B$. $B$ being countable means $F$ is countable too. Now given a set of basis vectors $\{\xi_1, \dots, \xi_n\} \in F$ every element of $\Q^n$ will give me a particular element of $\R$. But $|\Q| = |\Q^n| = \aleph_0$ so this would mean $|\R| = \aleph_0 \cdot \aleph_0 = \aleph_0$ which is false. Therefore $B$ must be uncountable.
Now what if I removed some elements from $B$? Suppose I removed $\kappa$-many basis vectors from $B$ where $\kappa \leq \aleph_0$. I’ll let $B’$ be what’s left in $B$ and $C$ will be the removed basis vectors so $B’$ and $C$ partition $B$ and $|C| = \kappa$.
Now let $G$ be the $\Q$-span of $B’$. I claim that $(G, +)$ is an uncountable subgroup of $(\R, +)$
Pf: first, $|B’| = \mathfrak c$ and $G\subsetneq \R$ so $\mathfrak c \leq |G| \leq \mathfrak c$ therefore $G$ is uncountable.
Next, consider $x,y \in G$. $x + y$ is still a finite combination of elements of $B’$ with rational coefficients so $x+y \in G$. Additionally, $0\in \Q$ means $0 \in G$ so $G$ has an identity, and if $x\in G$ then $-x \in G$ so $G$ has inverses. But because $B’$ is lacking some basis vectors $\R \backslash G \neq \emptyset$ which makes $G$ a strict subgroup of $\R$.
$\square$
Now I want to get a sense of what $\R / G$ looks like. Each coset is uncountable since they’re of the form $x + G$ for $x\in\R$. But an upper bound to the number of cosets that I can have is
$$
|\R / G| \leq | \R \backslash G| + 1
$$
since this would be the case if every set of the form $x + G$ for $x \in \R$ led to a unique coset. However $\R \backslash G$ is generated by $C$ which is at most countable, and this means the $\Q$-span of $C$ is also countable, so there are only countably-many cosets. Thus this was a whole lot of work to arrive at a partition of $\I$ no better than the very first one I warmed up with.
So this means I have to take $\kappa = \mathfrak c$ so $B’$ and $C$ are a partition of $B$ into two uncountable pieces. This means $G$ is still uncountable but I still need to determine if $\R / G$ is uncountable. This is how far I got before I came across Brian Scott’s extension of $\R / \Q$. I may return to this is a future post.