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So I have this question to prove a statement:

$O(n)\subset\Theta(n)$...

I don't need to know how to prove it, just that in my mind this makes no sense and I think it should rather be that $\Theta(n)\subset O(n)$.

My understanding is that $O(n)$ is the set of all functions who do no worse than $n$ while $\Theta(n)$ is the set of all functions that do no better and no worse than n.

Using this, I can think of the example of a constant function say $g(n)=c$. This function will surely be an element of $O(n)$ as it will do no worse than $n$ as $n$ approaches a sufficiently large number.

However, the same function $g$ would not be an element of $\Theta(n)$ as g does do better than $n$ for large $n$... Then since $g \in O(n)$ and $g \not\in \Theta(n)$, then $O(n)\not\in\Theta(n)$

So is the question perhaps wrong ? I've learnt it is dangerous to make that assumption and usually I have missed something, I just can't see what it might be in this case.

Any thoughts ? Thanks a lot..

Raphael
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Rawb
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2 Answers2

11

At the suggestion of Raphael, I have turned a previous comment into this answer.

It is not true that $O(f(n)) \subset \Theta(f(n))$. In fact, $\Theta(f(n)) = O(f(n)) \cap \Omega(f(n))$, by definition. So we have $\Theta(f(n)) \subset O(f(n))$.

Patrick87
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Think about it this way: every function that does "no worse than n" and "no better than n" is also a function that does "no worse than n". The "no better than n" part is just an additional constraint. This is a straightforward application of the logical rule that says: $x \wedge y \implies x$. By this reasoning, all functions that are in the set $\Theta(n)$ are also members of the set $O(n)$.

mrk
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