In general, what we may end up is something like this picture. We identify a set of nodes called the handle, some sets of nodes called teeth (like a comb), and we realise that the values on this set for all tours obey these rather complicated inequalities, which of course cut the previous fractional solution. These are called comb inequalities just because they resemble a comb. And they are very effective in cutting off more fractional inequalities. However even these inequalities are not enough, so that for some problems we have to resort to branching if we are not able to detect violating inequalities.
So we have to minimise the following Euclidean instance. First we get these three sub-tours. Then by adding sub-tour inequalities, we get this solution where the red lines are value one-half. We recognise here one comb inequality, then we recognise a second comb inequality and by adding these comb inequalities we get this new