# Is $(n^5 + n^7)\in \Omega(n^7)$? Shouldn't it be in $\Omega(n^5)$?

Is $$(n^5 + n^7)\in \Omega(n^7)$$? Shouldn't it be in $$\Omega(n^5)$$?

I understand Omega to be a "lower bound" on a function. Shouldn't the largest lower bound on the function $$n^5 + n^7$$ be $$n^5$$? (Just as the smallest upper bound is $$n^7$$)

The reason I say that the function is in the Omega class of functions larger than $$n^5$$ is because of the limit definitions of complexity [Source]: $$\require{enclose}$$

$$\begin{array}{c|c|c} \text{Big-}\mathcal O\ \text{notation} & \text{Comparison notation} & \text{Limit definition}\\ \hline f \in \mathcal o(g) & f \; {\scriptstyle \enclose{circle}{\kern .07em \bbox[6px] \lt \kern .07em}}\ g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} = 0\\ f \in \mathcal O(g) & f\; {\scriptstyle \enclose{circle}{\kern .07em \bbox[6px]\le \kern .07em}}\ g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} < \infty \\ f \in \Theta(g) & f\; {\scriptstyle \enclose{circle}{\bbox[7px]=}}\ g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} \in \mathbb{R}_{>0} \\ f \in \Omega(g) & f \;{\scriptstyle \enclose{circle}{\kern .07em \bbox[6px]\ge \kern .07em}}\ g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} > 0 \\ f \in \omega(g) & f\; {\scriptstyle \enclose{circle}{\kern .07em \bbox[6px]\ge \kern .07em}}\; g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} = \infty \\ \end{array}$$

Using the limit definition of $$\mathcal O$$, we correctly identify that $$n^5+n^7 \in \mathcal O(n^7)$$ and $$n^5+n^7 \notin \mathcal O(n^6)$$

However, using the limit definition of $$\Omega$$ tells us that the function is in $$\Omega(n^7)$$! But isn't the largest lower bound for this function $$n^5$$?

The function $$n^5 + n^7$$ is big-$$Ω$$ of both. However, it is little-$$\omega$$ of $$n^5$$.
It happens that $$n^5 + n^7 \in \Theta(n^7)$$. These asymptotic-notations only describe the behavior of the function $$f(n) = n^5 + n^7$$ for large $$n$$, aka "for all $$n > \textrm{some value}$$". The function is a polynomial function, and the highest-degree term determines the order of growth.
From the definition of $$Θ(g)$$, it means that function $$f$$ is both big-$$O$$ and big-$$Ω$$ of $$g$$. The function is bounded by some constant multiple of $$n^7$$ for large $$n$$. Namely, the function is bounded above by $$2n^7$$ and below by $$n^7$$.
Nevertheless, $$n^5 + n^7 \in \Omega{(n^5)}$$; the function is of [strictly] greater order than a quintic. Equivalently, $$n^5 \in O(n^5 + n^7)$$.
Observe that $$n^5+n^7 > n^7 \ge n^5, ~~\forall n \ge 1$$. Then clearly, the 'largest lower bound' is $$n^7$$. Therefore, $$n^5+n^7 = \Omega(n^7)$$ as well as $$n^5+n^7 = \Omega(n^5)$$ and also $$n^5+n^7 = \Omega(1)$$, but we generally choose the one that is largest and more tight.