There is a planar graph $G(V,E)$, for every $m$ the vertex set of the graph can be partitioned into $m$ sets $V_{1}$ .. $V_{m}$, such that for every $1 \leq i \leq m$, the graph formed on $V\setminus V_{i}$ has treewidth at most $3(m-1)$. Using the above Lemma how to show that every planar graph has a separator of size $2\sqrt{3n}+1$.
Ideas: as the partitions we can take the set of vertexes so the treewidth became $3(n-1)$ looks like it gives me nothing. The second approach we can orient the tree decomposition in a such way that the root of the the tree will be a separator for two subtree each one of them $\leq 2n/3$, this approach also goes nowhere. How to reach the result of separator of size $2\sqrt{3n}+1$
Let's develop the second idea using the assumption that $m=\sqrt{n}$. For a subtree of tree decomposition $T$, let $N(v)$ be the number of distinct vertexes in the labels of the subtree rooted at $v$. If $r$ is a root of initial tree decomposition $T$, then $N(r)=|V|$, where $V$ - vertexes of the graph $G$. Let $j$ be a tree node with $N(j) \geq 1/2$ and $N(u) \leq 1/2$ for all children $u$ of $j$. Let S, a separator be $X_{j}$, $|X_{j}| \leq 3(\sqrt{n}-1)$. Now let's make $j$ the root of $T$. So far so good, but what do you mean by $V_{i}$ is at most the average? Do separator $S$ and $V_{i}$ should be connected?