Proving $\Omega(cf) = \Omega(f)$
I think you're trying to take too many steps at once. Try to think about it one small step at a time.

Always think about:


What are the definitions I have available
What do I know already? What is proven?


Proving sets equal

Since you have two sets, $\Omega(cf)$ and $\Omega(f)$ and you want to prove them equal, you first need to think about what it means for two sets to be equal:

$$A = B \Longleftrightarrow A \subseteq B \land A \supseteq B$$

This tells you, that you need to show two things:


$\Omega(cf) \subseteq \Omega(f)$
$\Omega(cf) \supseteq \Omega(f)$


Proving $A$ to be a subset of $B$

For this, you need to apply the definition of $A \subseteq B$:

$$ A \subseteq B \Longleftrightarrow \forall a  \in A:(a \in B)$$

Nature of our proof

As you can probably see by now, our proof will have the following form:

$$\begin{align*}"\subseteq":\\ &amp; \text{Let } g \in \Omega(cf):\\
&amp;\Rightarrow \dots\\
&amp;\Rightarrow g \in \Omega(f)
\end{align*}
$$

The proof goes the same way for $"\supseteq"$.

Filling in the dots

Keeping this structure in mind, the path you need to follow is already laid out.

Using this definition for \Omega: $ \Omega(f) = \{ g |\, \exists\, n_0 \in \mathbb{N}, c &gt; 0 \forall n \geq n_0: cf(n) \leq g(n) \} $

One direction of the proof could look as follows:

$$\begin{align*}"\subseteq":\\ &amp; \text{Let } g \in \Omega(cf): \\
&amp;\Rightarrow \exists\, n_0\in \mathbb{N}, c'&gt;0: c'cf(n) \leq g(n)\ \ \forall n \geq n_0 &amp;\text{definition of } \Omega\\
&amp;\Rightarrow \exists\, d&gt;0,  n_0\in \mathbb{N}: d f(n) \leq g(n)\ \ \forall n \geq n_0 &amp;\text{Let } d=c'c\\
&amp;\Rightarrow g \in \Omega(f) &amp;\text{definition of } \Omega
\end{align*}
$$