# What do these Big-O notations mean in context of comparison

What do the following mean, in the context of greater than, or smaller than?

$$O(n \log ⁡n) > O(n)$$

$$O(nlogn) < O(n^2)$$

• Where you find such inequalities? Sep 17, 2020 at 11:00
• @zkutch, what do you mean by "inequalities" ? Sep 17, 2020 at 11:02
• I want to understand what: > or < between two Big-O notations for time complexity implies or means? Does it mean that the one that is greater is better, or the one that is less then is greater? Sep 17, 2020 at 11:05

In this context, this comparison means the subset of. Hence, $$O(n\log n) > O(n)$$ means All members of $$O(n)$$ exist in $$O(n \log n)$$ as well or $$O(n) \subset O(n \log n)$$. For example, $$f(n) = \sqrt{n} \in O(n)$$ and $$f(n) \in O(n\log n)$$ and $$g(n) = n \log n \not \in O(n)$$ and $$g(n) \ \in O(n \log n)$$.
• So, if, Algorithm1 is $$O(n)$$, and we modify it so, that it includes $$O(NlogN)$$, then we can exclude $$O(n)$$, and then the new algorithm is more dominant thus: $$O(n \log ⁡n) > O(n)$$ Sep 17, 2020 at 17:09
• So, what does this statement mean: $$O(nlogn) < O(n^2)$$ ?? Sep 17, 2020 at 17:11
• @JohnSmith You can interpret $O(n\log n) < O(n^2)$ such as the other case. For example, $n \sqrt{n} \in O(n^2)$, but it is not in $O(n\log n)$.