# Definitions > [!NOTE] Definition 1 (Big O of Real Sequence) > Let $f,g:\mathbb{N}\to \mathbb{R}$ be [[Real sequences|real sequences]]. Then $g:\mathbb{N}\to \mathbb{R}$ is a big O of, denoted $g(n)=\mathcal{O}(f(n)),$ if and only if, there exists $N\in \mathbb{N}$ and $c>0$ so that for all $n\geq N,$ $|g(n)|\leq c |f(n)|$ **Variations**: We may define $\mathcal{O}(f(n))$ to be the [[Sets|set]] of all functions satisfying the above condition and write $g(n)\in \mathcal{O}(f(n))$ instead of $g=\mathcal{O}(f(n)).$ Some sources may define as follows: $g(n)=\mathcal{O}(f(n))$ iff $\lim_{ n \to \infty } \frac{f(n)}{g(n)}\neq 0$ (or equivalently, $\lim_{ n \to \infty } \frac{g(n)}{f(n)}\neq \infty$) ; but this is more restrictive as discussed below. This definition may be useful for analysing monotone sequences. # Properties By [[Big O of Real Sequence in Terms of Limit]], if $\lim_{ n \to \infty } \frac{f(n)}{g(n)} \neq 0$ then $g(n)=\mathcal{O}(f(n))$ and conversely if $g(n)=\mathcal{O}(f(n))$ then $\frac{g(n)}{f(n)}$ is eventually bounded - not necessarily convergent (e.g $(-1)^{n}$ is bounded but not convergent). By [[Big Omega is Inverse Relation of Big O on Real Sequences]], $g(n)=\mathcal{O}(f(n))$ iff $f(n)=\Omega(g(n)).$ Note that we write $f(n)=\Omega(g(n)),$ pronounced $f(n)$ is [[Big Omega Relation on Real Sequences|big Omega]] of $g(n),$ iff there exists $N\in \mathbb{N}$ and $c>0$ so that for all $n>N,$ $|f(n)|\geq c |g(n)|.$ # Applications See [[Worst-case Running Time of Algorithm]].