# Trivial case of $s \in T$

I was reading paper A Formal Basis for the Heuristics Determination of Minimum Cost Paths in section B "Some Definition About Graphs" there is footnote that say "We exclude the trivial case of $s \in T$" I don't get it, is it trivial like empty set? or have different meaning? You can see the paper in here

Sorry, if this question not appropriate.

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By $\epsilon$, do you mean "Contained in"? –  Arthur Dec 30 '12 at 10:14
I'm not sure, but I guess yes, $s$ contained in $T$. It's an old paper from 1972, maybe the symbol for "element of" using the epsilon, I will change symbol. –  Kyomuu Dec 30 '12 at 10:17

This is what I can gather. We are starting from node $s$, and $T$ is a set of nodes we would like to reach with minimum cost. If $s\in T$, then we are already there, without any cost or search, so that particular case is not interesting in the least and trivial. The authors are therefore excluding such cases for the rest of the paper.
Ah! I get it, so there's no meaning to search, because $s$ is actually the goal node if $s \in T$, thanks for the answer. –  Kyomuu Dec 30 '12 at 10:36