For some time, I had the mis-belief that complex numbers are its own application domain, and that anything it can do can be achieved by applying complex arithmetic rules to real numbers.

However, when I scaned through the TOC for the second volume of my calculus textbook, I realized that complex number is a self-consistent in such way that it finds itself applications in a ubiquitous way.

However, instead of settling with it, I instead turn to look for generalizations of complex numbers that are as natural as generalization from real numbers to complex numbers, and that finds itself equal…

$$
\begin{align}
\int_0^1\left\lfloor\frac1x\right\rfloor^{-1}\,\rm {d}x
&=\sum_{k=1}^\infty\overbrace{\ \quad\frac1k\quad\ }^{\left\lfloor\frac1x\right\rfloor^{-1}}\overbrace{\left(\color{#C00}{\frac1k}-\color{#090}{\frac1{k+1}}\right)}^{\mathrm{d}x}\quad\leftarrow x\in\left[\frac1{k+1},\frac1k\right]\\
&=\sum_{k=1}^\infty\left(\color{#C00}{\frac1{k^2}}-\color{#090}{\left(\frac1k-\frac1{k+1}\right)}\right)\\
&=\color{#C00}{\sum_{k=1}^\infty\frac1{k^2}}-\color{#090}{\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)}\\

Suppose that B[a,b] is the metric space of all bounded real valued functions on [a,b], a<b with the sup metric. I want to prove that B[a,b] is NOT separable. Let $U=\{1_{\{t\}}: t\in [a,b]\}$. It is plain that U is uncountable. Let D be any dense subset of B[a,b]. Then there exists some $\phi_t\in D$ such that $d(1_{\{t\}}, \phi_t)=\sup | 1_{\{t\}}(x)- \phi_t(x)|<1/2$. Note that since $d(1_{\{t\}},1_{\{t'\}})=1$ for all $t\ne t'$, it follows that the map $g: U\to D$ defined by
$g(1_{\{t\}})= \phi_t$ is 1-1.