Let $G$ be a connected graph with vertices $V(G)$. A *bramble* of $G$ is a set of connected subgraphs $H\_1,\ldots,H\_n$ such that for each $i$ and $j$, $H\_i$ *touches* $H\_j$; that is, either $H\_i$ intersects $H\_j$ in a vertex, or there is an edge in $G$ that connects a vertex of $H\_i$ to a vertex of $H\_j$. The *order* of a bramble is the minimum size of a set $S\subset V(G)$ such that $S\cap H\_i$ is nonempty for all $i$. Then, the *bramble number* of $G$, denoted $Br(G)$, is the largest order of any bramble of $G$.


A *strict bramble* of $G$ is a set of connected subgraphs $H\_1,\ldots,H\_n$ such that for each $i$ and $j$, $H\_i$ intersects $H\_j$ in a vertex. The order of a strict bramble is similarly the minimum size of a set $S\subset V(G)$ such that $S\cap H\_i$ is nonempty for all $i$, and the strict bramble number of $G$, denoted $sBr(G)$, is the largest order of any strict bramble of $G$. (Strict brambles are sometimes called *intersecting families*, and the strict bramble number is sometimes called the “PI number" for “pairwise intersecting”.)


Since any strict bramble is also a bramble, we have $sBr(G)\leq Br(G)$ for all graphs $G$. My question is this: do we ever have $sBr(G)=Br(G)$? Or is it always the case that if there exists a strict bramble of order $k$, there exists a bramble of order $k+1$?


Edit: they are equal if G consists of one vertex and no edges. Are there any graphs with two or more vertices for which they are equal? (Thanks to Arun for pointing this out!)


(See [this post](https://mathoverflow.net/questions/142083/what-is-the-relationship-between-the-bramble-number-and-the-strict-bramble-numbe) for more discussion on bramble number and strict bramble number.)



Can the bramble number and the strict bramble number of a graph be equal?
Indeed, for every connected graph $G$ with at least two vertices, we have $sBr(G)<Br(G)$. This follows from a theorem by Seymour and Thomas and its proof (Theorem 12.4.3 in [Reinhard Diestel, Graph Theory, Springer GTM 173, 5th edition 2016](http://diestel-graph-theory.com/basic.html)). The theorem states that the treewidth of $G$ is equal to $Br(G)-1$.


Let $k=\mathrm{Br}(G)-1=\mathrm{tw}(G)$. By the assumption, $G$ has an edge, so it has a bramble of order $2$ and so $k\ge 1$. Let $(T,\mathcal{V})$ be a tree decomposition of $G$ of width $k$. This implies that for every edge $t\_1t\_2$ of the tree $T$, the intersection $V\_{t\_1}\cap V\_{t\_2}$ has at most $k$ vertices, since we can assume that the sets $V\_t \in \mathcal{V}$ are pairwise distinct.


Let $\mathcal{B}$ be a strict bramble in $G$. By the proof of Theorem 12.4.3, at least one of the following possibilities happens:


1) there is an edge $t\_1t\_2$ of $T$ such that $V\_{t\_1}\cap V\_{t\_2}$ covers $\mathcal{B}$: in this case the order of $\mathcal{B}$ is at most $k$.


2) there is a node $t$ of $T$ such that $V\_t$ covers $\mathcal{B}$, and we can assume that case 1) does not occur. We may assume that $|V\_t|=k+1$ and we want to show that a proper subset of $V\_t$ of $k$ vertices is sufficient to cover $\mathcal{B}$.


If $G$ has only $k+1$ vertices, then either $\mathcal{B}$ contains a singleton set, in which case $\mathcal{B}$ has order $1$, otherwise an arbitrary vertex of $V\_t$ can be omitted and the remaining $k$ still cover $\mathcal{B}$.


If $G$ has at least $k+2$ vertices, then $T$ has an edge $tt'$ such that $|V\_t\cap V\_{t'}|\ge 1$. By Lemma 12.3.1 in Diestel's book, $V\_t\cap V\_{t'}$ separates the sets of vertices of $G$ induced by the two components of $T-tt'$. 


I claim that $V\_t\setminus V\_{t'}$ covers $\mathcal{B}$. If not, there would be a set $B' \in \mathcal{B}$ disjoint with $V\_t\setminus V\_{t'}$, but also a set $B \in \mathcal{B}$ disjoint with $V\_t\cap V\_{t'}$ (by the assumption that $V\_t\cap V\_{t'}$ does not cover $\mathcal{B}$). By the previous paragraph, this means that $B$ and $B'$ are disjoint, which contradicts the strictness of $\mathcal{B}$.


