# Which one of the following is the correct asymptotic notation? [closed]

While studying the complexity theory, I encountered a question which is as follows:-

Which one of the following is correct?

1) θ(g(n)) = O(g(n)) ∩ Ω(g(n))

2) θ(g(n)) = O(g(n)) ∪ Ω(g(n))

I know what, θ and Ω represent. But, I am totally confused with the union and intersection operations performed between them. What does it actually mean and what is the correct answer? Thanks for the help!

## closed as off-topic by Emil Jeřábek, Gamow, Sasho Nikolov, Hsien-Chih Chang 張顯之, AryehDec 29 '18 at 22:08

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We can think of $$\mathcal O(g(n))$$ as being a family of functions: those which increase either slower than or as fast as $$g(n)$$. For example, if $$g(n) = n^2$$ then the function $$f(n) = n$$ satisfies $$f(n) \in \mathcal O(g(n))$$.
Similarly, $$\Omega(g(n))$$ is also a function family of the functions which increase either faster or as fast as $$g(n)$$. Again if $$g(n) = n^2$$ then $$f(n) = n^3$$ satisfies $$f(n) \in \Omega(g(n))$$.
Now $$\Theta(g(n))$$ is (yet another) function family of the ones which increase as fast as $$g(n)$$ (but not faster or slower!) For example $$g(n) = n^2$$ and $$f(n) = 2n^2+3n+1$$ satisfy $$f(n)\in \Theta(g(n))$$.
If $$\Theta(g(n)) = \mathcal O(g(n)) \cup \Omega(g(n))$$, then we would have that $$f(n) = n \in \Theta(g(n))$$ for the same $$g(n)$$ as before, which is not true. On the other hand, $$\Theta(g(n)) = \mathcal O(g(n)) \cap \Omega(g(n))$$ would imply that $$\Theta(g(n))$$ only contains functions which increase at the same speed as $$g(n)$$, which is what we want.