Let's check together whether the TSP-decision problem is NP-complete. Maybe it will help me to understand things better. Question for TSP-decision problem: Given n cities and a tour from length k. The traveller starts at an arbitrary city, visits every other city just once and returns to the starting point. Does a tour from length <= k exist? Let's first check whether the TSP is in NP: A proof is given. Proof in this case is a tour. For it to be in NP, we must be able to verify this proof with a deterministic algorithm in polynomial time. So first we have to check, whether every city is only visited once. This can be done at most in O(n^2). Next we need to calculate the distances and sum them up. This can be done as well at most in O(n^2). The last step is to check whether the calculated length is <= k. The hole process would require a polynomial time --> 2 * n^2 = O(n^2) So the TSP-decision problem is in NP. What about the NP-hardness? We don't need to prove that, because Richard M. Karp proved that the Hamiltonian Circuit is NP-complete. The TSP is a special case of the Hamiltonian Circuit, therefore we know TSP must be NP-complete as well. NP-complete means = NP-hard and NP. Is this possible with TSP-optimization problem? Question for TSP-optimization: Given n cities and a tour from length k. A traveller starts at an arbitrary city and visits every other city just once and returns to the starting point. Is tour from length k the shortest tour? We can check in polynomial time, that every city is visited once. However we can't check whether k is really the shortest tour, because we would have to check every other possible tour as well. That would mean (n-1)! possible tours. And this would make the hole process above exponential. So TSP-optimization problem is not in NP !? Therefore it is not NP-complete, but NP-hard !?Anything wrong about my thoughts? 
TSP decision problem vs TSP optimization problem
complexity theory;np complete;np hard;np;traveling salesman