Contructing a $\delta$-fine tagged partition from the old ones

Question

Let $[a,b]\subset \mathbb{R}$. A tagged partition of $[a,b]$ is a set $D=\{(t_i,I_i)\}_{i=1}^m$ where $\{I_i\}_{i=1}^m$ is a partition of $[a,b]$ consisting of closed non-overlapping subintervals of $[a,b]$ and $t_i\in I_i$; $t_i$ is called the tag associated with $I_i$.

Suppose that $D=\{(t_i,I_i)\}_{i=1}^m$ and $D'=\{(s_j,J_j)\}_{i=1}^n$ are any two tagged partitions of $[a,b]$. We say that $D'$ is finer than $D$ if for each $j\in\{1,2,\cdots,n\}$, there exists $i\in\{1,2,\cdots,m\}$ such that $J_j\subset I_i$ and every tag in $D$ is a tag in $D'$.

Let $\delta:[a,b]\to (0,+\infty)$ be a positive function. We say that a tagged partition $D=\{(t_i,I_i)\}_{i=1}^m$ is $\delta$-fine if $$I_i\subset(t_i-\delta(t_i),t_i+\delta(t_i))\quad\mbox{for } i=1,2,\cdots,m.$$

Question:

Suppose $D$ and $D'$ are $\delta$-fine tagged partitions of $[a,b]$. How do we construct a third $\delta$-fine tagged partition $E$ of $[a,b]$ such that $E$ is both finer than $D$ and $D'$?

Answer 1

There are two things to keep track of:

• The partitions;
• The tags.

On an intuitive level, they can be taken care of by the family $(I_i \cap J_j)_{i,j}$, placing the old tags as appropriate, and making up some new ones for those intervals not containing an old tag (this may require further splitting of the intervals, depending on $\delta$). We also need to split those $I_i \cap J_j$ that contain both $t_i$ and $s_j$ (presuming they are distinct, of course).

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A more detailed version of the sketch above:

Denote $K_{ij} = I_i \cap J_j$; then $(K_{ij})_{ij}$ is a partition of $[a,b]$ of the required form.

Note that if we produce a tagged partition $E_{ij}$ for each $K_{ij}$, we can amalgamate these into a tagged partition $E$ of $[a,b]$. So fix a $K_{ij}$, and distinguish four cases:

• $t_i, s_j \in K_{ij}$: if $t_i = s_j$, use the tagged partition $\{(K_{ij},t_i)\}$. Otherwise, use the partition $\{(K_{ij}\cap[a,r],t_i \wedge s_j),(K_{ij}\cap[r,b],t_i \vee s_j)\}$, where $r = \frac12(t_i+s_j)$, and $\wedge$ and $\vee$ denote the binary $\max$ and $\min$ operations, respectively.
• $t_i \in K_{ij}$, $s_j \notin K_{ij}$: use the tagged partition $\{(K_{ij}, t_i)\}$.
• $t_i \notin K_{ij}$, $s_j \in K_{ij}$: use the tagged partition $\{(K_{ij}, s_j)\}$.
• $t_i, s_j \notin K_{ij}$: For each $r \in K_{ij}$, let $L_r = (r-\frac12\delta(r),r+\frac12\delta(r)) \cap K_{ij}$. Then each $L_r$ is open in the subspace topology, and $\bigcup\limits_{r \in K_{ij}} L_r = K_{ij}$. Since $K_{ij}$ is compact, find a finite subset $r_k, 1\le k \le \ell$ such that $\bigcup_k L_{r_k} = K_{ij}$ and $k<k'$ implies $r_k < r_{k'}$. Denote with $\bar L_r$ the closed interval corresponding to $L_r$. Use the tagged partition $$\{(\bar L_{r_k}, r_k):1\le k\le\ell\}$$ where $L_{r_0} = \varnothing$ (we will let rest the boring details for dealing with the non-trivial overlapping of $L_{r_k}$).

This construction yields a tagged partition $E_{ij}$ of each $K_{ij}$, and hence of $[a,b]$; call this latter tagged partition $E$. It is to be verified that $E$ refines $D$ and $D'$.

Now each $K_{ij}$ is contained in both $I_i$ and $J_j$, and hence all intervals used for $E_{ij}$ are, too; furthermore, each tag $t_i, s_j$ occurs in some $K_{ij}$, and by construction remains a tag in $E$. Thus $E$ refines $D$ and $D'$.

Finally, it remains to establish that $E$ is $\delta$-fine. We restrict attention to the tags within one $E_{ij}$. If this $E_{ij}$ was formed using one of the first three constructions, it is immediate from the $\delta$-fineness of $D$ and $D'$ that $E_{ij}$ is $\delta$-fine. For the fourth construction, observe that $L_{r_k}$ is strictly contained in $(r_k-\delta(r_k),r_k+\delta(r_k))$, since $\delta(r_k)>0$; therefore, $\bar L_{r_k}$ is also contained in this set, and hence so are the intervals used in the partition. In conclusion, all $E_{ij}$, and consequently $E$, are $\delta$-fine.