Do any decision problems exist outside NP and NP-Hard?

If NP-hard problems are those which are at least as hard as those in NP, then shouldn't NP lie completely in NP-Hard?

All Problems in NP can be mapped to itself, implying that all NP problems are NP-hard problems. So, NP-hard should completely cover NP. I know I am surely wrong somewhere. I would like to know what is wrong with this argument.
Answer:

The proposition is: "NP- hard problems are at least as hard as the HARDEST problem in NP." This Hardest part is NP complete problems. And these hardest problems (NPC) can be reduced to each other. And all those problems which are NP hard say halting problem, and it does not lie on NP. So NP does not lie completely in NP-Hard.

Sumit Kushwaha at Quora Visit the source

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Other answers

NP-Hard problems are divided into two parts: Problems for which we know desired output but there is no algorithm through which we can achieve it. Such problems are not in NP. Problems for which there exists a non-deterministic algorithm (executable in polynomial time). Such problem lies in NP and known as NP-Complete.

Jinal Dhruv

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