In a topological space, if we have two topologically distinct points $a,b\in X$, can we always use two nets $(x_i)_{i\in I}$ and $(y_i)_{i\in I}$ with a same index set $I$ to approach each point, respectively? I've never thought about this subtlety before. But I just realized, in a topological vector space, this is always true because vector translation is a homeomorphism, and so the poset of neighborhoods at each point is exactly the same everywhere. So that guarantees two nets approaching two points share the same poset as their index set. But can this be always true in general topological spaces?
Hope this make sense. But honestly I don't know if it does. Thanks in advance.
Yes: given $x^1_i\stackrel{i\in I_1}\longrightarrow L_1$ and $ x^2_i\stackrel{i\in I_2}\longrightarrow L_2$, consider $(I_1\times I_2,\preceq)$ with $(i_1,i_2)\preceq (i_1',i_2')\Leftrightarrow i_1\preceq_1 i_1'\land i_2\preceq_2 i_2'$ and define $\tilde x^{1}_{i_1,i_2}=x^1_{i_1}$ and $\tilde x^2_{i_1,i_2}=x^2_{i_2}$. Consider the nets $(\tilde x^j_{i_1,i_2})_{(i_1,i_2)\in I_1\times I_2}$: I claim that for all $j$, $$\tilde x^j_{i_1,i_2}\stackrel{(i_1,i_2)\in I_1\times I_2}\longrightarrow L_j$$ In fact, given any $U^j\in \tau$ such that $U^j\ni L_j$ ($j=1,2$), there is some $h_j$ such that $x^j_i\in U^j$ for all $i\succeq_j h_j$. Now, let $(i_1,i_2)\in I_1\times I_2$ and consider $v=\begin{cases}(h_1,i_2)&\text{if }j=1\\ (i_1,h_2)&\text{if }j=2\end{cases}$ (i.e. $v$ has the same entries as $(i_1,i_2)$, except for the $j$-th, where it is equal to $h_j$). It's quite clear that $\tilde x^j_{s_1,s_2}\in U^j$ for all $(s_1,s_2)\succeq v$.
Notice that we have actually proved that $\left((\tilde x^1_{s},\tilde x^2_{s})\right)_{s\in I_1\times I_2}\to (L_1,L_2)$ in the product topology of $X\times X$.
The same can be replicated word by word with any family of $x_i^r\stackrel{i\in I_r}\longrightarrow L^r$, $r\in R$, under the assumption that $\prod_{r\in I}I_r\ne\emptyset$. The new indexing set will be $\prod_{r\in R}I_r$ with $f\preceq g\Leftrightarrow\forall r\in R, f_r\preceq_r g_r$, and the new nets are $(\tilde x^r_f)_{f\in\prod_{r\in\Bbb R}I_r}$ with $\tilde x^r_f=x^r_{f_r}$.
Just like before, $(\tilde x^\bullet_f)_{f\in\prod_{r\in R}I_r}$ converges to $L_\bullet$ in the product topology of $X^R$. With the axiom of choice, we have the stronger result that the net $(\tilde x^\bullet_f)_{f\in \prod_{r\in R} I_r}$ converges to $L_\bullet$ in the box topology of $X^R$.