This was written in Page 92, of Higson's Analytic $K$-Homology book.
Let $H$ be a hilbert space. The $C^*$ algebra $K(H)$ of compact operators is the direct limit of a sequence $$M_2(\Bbb C) \subseteq M_4(\Bbb C) \subseteq \cdots $$
How is this so? I suppose that $H$ is an infinite dimensional $\Bbb C$ vector space.
I know any compact operator is the limits (in operator norm) of finite rank operator. But this still doesn't really explain the direct limit...
Also why are we only considering matrices of size $2^n$?
The direct limit of the sequence $M_2(\mathbb{C}) \subset M_4(\mathbb{C}) \subset \cdots$ with diagonal maps $a \mapsto \left(\begin{array}{cc} a & 0 \\ 0 & a\end{array} \right)$ will give you the $2^{\infty}$ UHF algebra (CAR algebra), not the compact operators. Perhaps you mean $a \mapsto \left(\begin{array}{cc} a & 0 \\ 0 & 0\end{array} \right)$? In that case everything in the limit is the limit of finite rank operators so must be contained in $\mathcal{K}$.
On the other hand, observe that you can get any finite rank operator in the limit since, for any $n$, you can find $k$ such that $M_n(\mathbb{C}) \subset M_{2^k}(\mathbb{C})$, so in fact the limit must be $\mathcal{K}$.