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I'm working on a homework problem that asks to prove that $Θ(n)+O(n^2 )≠O(n^2 )$.

I'm not looking for an answer to the problem, just a better understanding of what it is that needs to be proved.

The previous question in the homework had me prove $Θ(n)+O(n^2 )⊆O(n^2 )$, which I feel I have successfully done. So how does one go from proving that $Θ(n)+O(n^2 )$ is a subset of $O(n^2 )$ to proving $Θ(n)+O(n^2 )≠O(n^2 )$?

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    The title is the problem you already solved, you should fix that. – 2017-01-28
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    Actually, $Θ(n)+O(n^2 )=O(n^2 )$, please see comment on the answer below. – 2017-01-28

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You must show that $O(n^2)$ is not a subset of $\Theta(n) + O(n^2)$. So you need to find a function that is $O(n^2)$ but not $\Theta(n) + O(n^2).$ A helpful thing to remember is that in $O(n^2),$ $n^2$ is allowed to overestimate the growth (it is an upper bound), but not in the case of $\Theta(n).$ If $f(n)\in\Theta(n),$ $n$ must be its true asymptotic growth.

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    Yes they would have to do that to prove this result -- but this result does not hold: for every $f(n)$ in $O(n^2)$, since $n$ is in $\Theta(n)$ and in $O(n^2)$, $f(n)-n$ is in $O(n^2)$ hence the identity $f(n)=n+(f(n)-n)$ shows that $f(n)$ is in $\Theta(n)+O(n^2)$. – 2017-01-28
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    @Did I noticed that too but assumed we were thinking like computer scientists, not mathematicians. – 2017-01-28
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    @Did (i.e. all functions must be positive) – 2017-01-28