Is it true that O(f(n)) is less than or equal to Ω(f(n))?

Let's break it down with definitions: f(x) = O(g(x)) means |f(x)| ≤ M * g(x) for some constant M > 0, while f(x) = Ω(g(x)) means |f(x)| ≥ M * g(x). Given this, your statement holds true for cases where |f(n)| = M * g(n). If it were a strict inequality, though, then it wouldn’t be valid anymore. Also keep in mind that there's another notation, Θ(f(n)), which describes functions that fit both O and Ω.

Big O represents the worst-case performance, while big Omega describes the lower bound. So saying O(f(n)) is greater than or equal to Ω(f(n)) isn’t quite right. Both big O and big Omega refer to growth behavior of functions, regardless of context. It can be confusing since they don't correlate directly with best or worst case scenarios like we often assume.

Can someone just explain this? I'm still trying to wrap my head around it.

So, applying the less than or equal sign in this context doesn’t really make sense. When we say f(x) = O(g(x)), we’re saying that f(x) is part of the set that O(g(x)) describes. They're not equal, and comparing sets directly with inequalities isn't valid. If you’re thinking in terms of subsets, then they don’t overlap completely either. For example, take f(n) = n; O(n) includes constant functions, which aren't part of Ω(n). So, the statement is pretty questionable.