I'm reading a proof of this claim, and there is a claim that if $\tau_n=\tau+\frac{1}{2^n}$ where $\tau$ is a stopping time, then we can write the stochastic interval $[0,\tau_n]$ as
$[0,\tau_n]=(\{0\}\times\Omega)\cup(\bigcup_{k\in\mathbb{N}_0}(\frac{k}{2^n},\frac{k+1}{2^n}]\times \{\tau\ge\frac{k}{2^n}\}).$
I'm having trouble seeing $LHS\subset RHS$ when $t>0$. Suppose $(t,\omega)\in[0,\tau_n]$, i.e. $0< t\le\tau(\omega)+\frac{1}{2^n}$. Then $t\in(\frac{k}{2^n},\frac{k+1}{2^n}]$ for some $k\in\mathbb{N}_0$, so $t-\frac{1}{2^n}\in(\frac{k-1}{2^n},\frac{k}{2^n}]$. Then we have $\tau(\omega)\ge t-\frac{1}{2^n}$, which does not preclude $\tau(\omega)<\frac{k}{2^n}$ (e.g. $t-\frac{1}{2^n}=\frac{k-1/2}{2^n}<\tau(\omega)<\frac{k}{2^n}$ seems possible).
I tried repairing this issue by changing the product to $[0,\tau_n]=(\{0\}\times\Omega)\cup(\bigcup_{k\in\mathbb{N}_0}(\frac{k}{2^n},\frac{k+1}{2^n}]\times \{\tau\ge\frac{k-1}{2^n}\}),$ but this creates an issue in showing $RHS\subset LHS$.
Would greatly appreciate some input. Thanks in advance!
Saz and Did point out that I was misinterpreting the notation describing $\tau_n$. With the correct interpretation, the proof is as follows:
$[\subset]:$ Pick $(t,\omega)\in[0,\tau_n]$, i.e. $0\le t\le \tau_n(\omega)$. Note that if $t=0$, then $(0,\omega)\in\{0\}\times\Omega.$ If $t>0$, then $t\in(\frac{k}{2^n},\frac{k+1}{2^n}]$ for some $k\in\mathbb{N}_0$. So $0\le\frac{k}{2^n}<t\le\tau_n(\omega)$, which implies $k<2^n\tau_n(\omega)$, which cannot happen if $\tau(\omega)<\frac{k}{2^n}$, i.e. $\tau(\omega)\ge\frac{k}{2^n}$. Hence $(t,\omega)\in\cup_{k\in\mathbb{N}_0}(\frac{k}{2^n},\frac{k+1}{2^n}]\times \{\tau\ge\frac{k}{2^n}\})$.
$[\supset]:$ Pick $(t,\omega)\in RHS$. If $t=0$, $(t,\omega)\in[0,\tau_n]$, since $\tau_n\ge0$. If $t>0$, then $t\in(\frac{k}{2^n},\frac{k+1}{2^n}]$ for some $k\in\mathbb{N}_0$ and $\tau\ge\frac{k}{2^n}$; the latter implies $\tau_n\ge\frac{k+1}{2^n}\ge t$. So $(t,\omega)\in[0,\tau_n]$.