I was reading “Introduction to Algorithms” by CLRS and it says
Note that f(n) = Θ(g(n)) implies f(n) = O(g(n)) since Θ notation is a stronger notation than O notation. Written set theoretically, we have Θ(g(n)) ⊆ O (g(n)) .
This means that Θ is a subset of O . Both the statements seem contradictory.
Lets refactor and reword these statements for ease of thought.
Let $A(n)$ be $Θ(g(n))$.
Let $B(n)$ be $O(g(n))$.
Note that $A$ implies $B$ because $A$ is stronger than $B$.
This means for $A$ to be fulfilled we have to fulfill the criteria of $B$ and more. Therefore $A$ being fulfilled implies $B$ must also be.
Written set theoretically, we have the set of functions that fulfill the criteria of $A$ are a subset of the functions that fulfill the criteria of $B$.
Well this is not contradictory, but actually equivalent in a sense to the last statement which showed that $A$ has the criteria of $B$ and more.
From our understanding of this point we can determine that: All the functions that fulfill the criteria of $A$ fulfill $B$, but not all functions that fulfill the criteria of $B$ fulfill the criteria of $A$.
This is really an equivalent statement to: The functions that fulfill the criteria of $A$ are a subset of the functions that fulfill the criteria of $B$ because the second set contains the first.
And therefore $\Theta(g(n))\subseteq O (g(n))$ is true, and it is equally true that $f(n) = \Theta(g(n))$ implies $f(n) = O(g(n))$ without any contradiction.
Furthermore this makes sense with the definition for $\Theta(g(n))$ which states that: $f(n)\in \Theta(g(n))$ if and only if $f(n) \in O(g(n))$ and $f(n) \in \Omega(g(n))$. Clearly by this definition, $f(n) \in Θ(g(n))$ does imply $f(n) \in O(g(n))$, and $\Theta(g(n))$ is clearly a subset of $O(g(n))$ as well.
No contradiction here. For simplicity, consider this example. You have a function, $f(n)$, if we say $f$ is $O(n^2)$ that would mean that $f$ is asymptotically less than a quadratic function. But if we know that $f$ is in fact linear, that would imply even stronger statement that $f$ is actually $\Theta(n)$. But linear functions are obviously one of the many functions bounded by quadratic function (such as $\log$, $n^{3/2}$, etc), hence it is a subset.
