Big Omega/Oh Notation (Application)

Question

For clarity; I use the following definitions (taken from Wikipedia), in my question:

Big Omicron: $f(n)=O(g(n))$. Formal definition: $\exists k >0,\exists n_{0}, \forall n>n_{0}: |f(n)| \leq k |g(n)| $.

Big Omega: $f(n)=\Omega(g(n))$. Formal definition: $\exists k >0,\exists n_{0}, \forall n>n_{0}: f(n) \geq k g(n) $.

Property 1: $f(n)=O(g(n)) \Leftrightarrow g(n)=\Omega(f(n))$

Property 2: $f(n)=\Omega(g(n))$ and $g(n)=\Omega(h(n))$, then $f(n)=\Omega(h(n))$

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Now consider the following: $v(n)=v_{1}(n)+v_{2}(n)+v_{3}(n)$. $v_{1}(n)$ and $v_{2}(n)$ are positive for all $n>0$. $v_{3}(n)$ is negative for all $n>0$. It is also true that $v_{1}(n)+v_{2}(n)+v_{3}(n) \geq 0$ for all $n>0$.

Using the definitions stated above, is the following correct:

$v_{1}(n)+v_{2}(n)=O(v_{1}(n)+v_{2}(n)+v_{3}(n))$ because $|v_{1}(n)+v_{2}(n)| \leq k|v_{1}(n)+v_{2}(n)+v_{3}(n)|$

and, because of property 1, we can write

$v_{1}(n)+v_{2}(n)+v_{3}(n)=\Omega(v_{1}(n)+v_{2}(n))$

and because of property 2 it holds that

$v(n)=v_{1}(n)+v_{2}(n)+v_{3}(n)=\Omega(v_{1}(n))$

My guess is that something is wrong because it feels odd to lower-bound the function $v(n)$ without considering the negative term $v_{3}(n)$.

Any help is much appreciated.

Answer 1

$v_{1}(n)+v_{2}(n)=O(v_{1}(n)+v_{2}(n)+v_{3}(n))$ because $|v_{1}(n)+v_{2}(n)| \leq k|v_{1}(n)+v_{2}(n)+v_{3}(n)|$

This is false. Where did you get it from?