TITLE: NP not equal to SPACE(n)
QUESTION [10 upvotes]: Exercise 3.2 of Computational Complexity, a Modern Approach states:
Prove: NP != SPACE(n) [Hint: we don't know if either is a subset of the other.]
I don't know how to solve this problem.
It's in the diagonalization chapter.
I've looked around google a bit, but it basically ends up linking back to the Arora/Barak book.
Anyone know how to attack this?
Thanks!
More generally: to prove a language to be uncommputable, I can use diagonalization -- but to prove that two sets of languages (Space(N) and NP) are different, when it's not known that either is contained in the other -- what techniques are there for these proofs?
Thanks again!

REPLY [12 votes]: I think that a common technique for proving such statements is for example the following type:
One class shares a closure property, while the other cannot because of a hierarchy theorem.
Thus they cannot be equal.
In this particular case a proof could proceed along these lines: Since NP is closed under polynomial time reductions, so would SPACE(n), if they were equal. Then deduce that polynomial time reductions would imply that SPACE(n^2) is contained in SPACE(n), which is impossible by the space hierarchy theorem.

REPLY [8 votes]: Scott Aaronson has a blog post, Sidesplitting Proofs, which is a highly recommended read for a variety of reasons. The first proof in the list, which is said to be folklore, is that E, the class of problems solvable in $2^{O(n)}$ time, is not equal to PSPACE. The key is again padding: if the two are equal, then E=EXP, and we derive a contradiction. Like in your case, which is bigger (if one is even contained in the other) is not known.