exists $w \in{\cal V}$ such that $$d(w,u)=1,$$ and $$d(w,v)=d(v,u)-1,$$ i.e., $w$ is a neighbor of $u$ and is on the shortest path between $u$ and $v.$ Note in a tree structure $w$ is unique.

If ${\cal I}\cap T^{-w}_u=\emptyset,$ then $\forall a \in {\cal I},$ $$d(w,a)=d(u,a)-1<d(u,a),$$ which contradicts the fact that $u$ is a Jordan infection center.

If ${\cal I}\cap T^{-w}_u\neq \emptyset.$ Since $\forall b \in {\cal
I}\cap T^{-w}_u,$ $$d(v,b)=d(v,w)+d(w,b),$$ i.e., $$d(w,b)=d(v,b)-d(v,w)\leq
\lambda-1.$$ On the other hand, since $\tilde{e}(u)=\lambda,$ $\forall h \in T^{-u}_w\cap {\cal I},$ $$d(w,h)=d(u,h)-1\leq
\lambda-1.$$ In a summary, $\forall h \in {\cal I},$ $$d(w,h)\leq
\lambda-1,$$ which contradicts the fact that the minimum infection eccentricity is $\lambda.$

Therefore all Jordan infection centers must be adjacent to each other. However, suppose there exist $n$ infection eccentricity centers where $n>2$, they would form a clique with $n$ nodes which contradicts the fact that the graph is a tree. Therefore, there exist at most two adjacent Jordan infection centers.

Based on Lemma \[lem:mainlemma\] and Lemma \[lem:numofcenters\], we finish this section with the following theorem.

\[thm:OSPD source\] Consider a tree network with infinitely many levels. Assume that the observed infection topology ${\bf Y}$ contains at least one infected node. Then the source node associated 