Attempt: Suppose, that for some $c \in A$, it happens. At least for two intervals $I_s$ and $I_t$, it happens. For this to occur, the intervals must lie on two "opposite sides" of $c$ since $c \in A$ and $c \notin \bigcup I_n$.
Now, any of the following pairs hold: $\{r_s= x_s-e, r_t=x_t-e\};\{r_s= x_s-e, r_t=\pi-x_t\};\{r_s=\pi- x_s, r_t=\pi -x_t\};\{r_s= \pi-x_s, r_t=x_t-e\} $.
We must have $\displaystyle x_s-\frac{r_s}{2^{100s}}=x_t+\frac{r_t}{2^{100t}}$. But it is not possible in any case since $x_s-x_t$ is rational and $\displaystyle\frac{r_s}{2^{100s}}+\frac{r_t}{2^{100t}}$ is irra…
Prove that the empty relation, defined on a non-empty set $A$, is asymmetric.
My work:
We need to show that if $\mathcal{R}=\varnothing$ is a relation on $A\times A$, with $A\neq\varnothing$, then for all $x,y\in A$, if $(x,y)\in\mathcal{R}$ then $(y,x)\notin\mathcal{R}$.
Proof Since $A\ne...
