why the big-oh of a linear function is n^2?

Question

I know the definitions and differences of big-theta, big-oh, and big-omega in finding time complexity of algorithms. what I can't understand is that why the big-oh of a linear function cannot be O(n).

$0 \le an + b \le cn \Rightarrow c = a + \frac b {n_0}$

so we could find a value for constant $c$. as a result, the big-oh of a linear function can also be $n$. am I right??

Note: I have read all the previous answers and questions in this commission and others. but I didn't understand. that's why asked a new question

Answer 1

A function does not have a unique big-oh. $O(f(n))$ denotes a class of functions that are big-oh of $f(n)$. Big-oh is used to give upper bounds.

For example, $f(n)=3n+5$ is $O(n)$, but it is also $O(n^2)$ and also $O(n^3)$ and also $O(2^{2^{2^n}})$. However, $f(n)$ is (in this example) not $O(\log n)$.

In set theory notation, we have $O(n)\subset O(n^2)$, i.e. any function that is $O(n)$ is also $O(n^2)$. However, if we had a function that is $O(n)$ and you gave $O(n^2)$ as an upper bound, that would not be incorrect, but it would be a weaker result than saying it is $O(n)$.

$\Omega$ works the other way around (i.e., it is used for lower bounds). A function that is $\Omega(n^2)$ is also $\Omega(n)$. $\Theta$ is used to give "tight" bounds: a function that is $\Theta(n^2)$ is neither $\Theta(n)$ nor $\Theta(n^3)$.