Wikipedia states that “On infinite graphs with a finite branching factor and edge costs that are bounded away from zero $ (d(x,y)>\varepsilon >0$ for some fixed $ \varepsilon$ ), A* is guaranteed to terminate only if there exists a solution.”

Does this mean that, if I have such a graph, then there is no admissible A* which exists for this graph? When one says that A* is not admissible, this means that its heuristic is not admissible, right?

Furthermore, is it correct to say that a heuristic is only admissible with respect to a graph – not generally admissible?

For example, if I have a infinite graph which has a finite branching factor, and the cost of each edge is half of the cost of the edge preceding it (something like this: $ goal\leftarrow_{_{c=2}} start \rightarrow_{_{c=1}}q1 \rightarrow_{_{c=1/2}} q_2 \rightarrow …$ ), the heuristic and therefore the A* is necessarily inadmissible as there exists no fixed $ \epsilon>0$ that is less than the cost of any edge?

To generalize, the $ epsilon$ constraint is to make sure that there is no infinite path who’s total cost converges, thereby assuring termination?

Any clarification is appreciated. Thanks!