3

The same question has been asked here

Some authors define $\Omega$ in a slightly different way: let’s use $ \overset{\infty}{\Omega}$ (read “omega infinity”) for this alternative definition. We say that $f(n) = \overset{\infty}{\Omega}(g(n))$ if there exists a positive constant $c$ such that $f(n) \geq c\cdot g(n) \geq 0$ for infinitely many integers $n$, whereas the usual $\Omega$ requires that this holds for all integers greater than a certain $n_0$.

Show that for any two functions $f(n)$ and $g(n)$ that are asymptotically nonnegative, either $f(n) = O(g(n))$ or $f(n)= \overset{\infty}{\Omega}(g(n))$ or both, whereas this is not true if we use $\Omega$ in place of $\overset{\infty}{\Omega}$

The solution given in the link hints that we can negate the definition of $ \overset{\infty}{\Omega}$ which gives us the definition of big O.

But negation of the definition of $ {\Omega}$ also gives us the same result, doesn't it?

I cannot wrap my head on how the condition of infinitely many integers makes difference rather than saying for all n greater than certain $n_0$.

Please help me understand this.

Community
  • 1
user422489
  • 43
  • 4
  • if n is odd then ​ f(n) = 1 ​ else ​ f(n) = n ​ ​ ​ –  Mar 06 '17 at 16:49
  • 1
    If the two definitions, once negated, become equivalent, they were already equivalent before the negation. Since that does not hold, you probably made some mistake when negating the definition of $\Omega$. Be very careful when negating quantifiers: in my experience, many fail at that point. – chi Mar 06 '17 at 18:17
  • I need help in negating the definition of ${\Omega}$. Negation of infinitely many integers is that there are fixed number of integers of which there will be some maximum integer say ${n_0}$ after which condition won't hold true. Negation of n greater than ${n_0}$ is again there is some maximum ${n_0}$ after which condition won't hold true. I reach the same conclusion in both cases. – user422489 Mar 06 '17 at 18:24
  • Consider the set of even positive integers. ​ ​ –  Mar 07 '17 at 04:47

1 Answers1

2

When you negate $\Omega$ you get $\stackrel{\infty}{O}$, the analog of $\stackrel{\infty}{\Omega}$. I suggest writing the definition of $\Omega$ as a first-order formula and then negating it.

To see the difference between $\Omega$ and $\stackrel\infty\Omega$, consider the functions $f(n) = n$ and $$g(n) = \begin{cases} 1 & \text{$n$ even,} \\ n^2 & \text{$n$ odd.} \end{cases}$$ You can check that $f(n) \neq O(g(n))$ and $f(n) \neq \Omega(g(n))$, whereas $f(n) = \stackrel\infty O(g(n))$ and $f(n) = \stackrel\infty\Omega(g(n))$.

Yuval Filmus
  • 276,994
  • 27
  • 311
  • 503
  • Got it. I was not thinking about cases where f(n) won't belong to both big Oh and big Omega. That example helped. Thanks – user422489 Mar 19 '17 at 13:36