Problem 11D in Willard's General topology textbook states that:
Diagonal principle) If $(x_i)$ converges to $x$ and, for each $i\in I$, a net $(x^i_j)_{j\in J_i}$ converges to $x_i$, then there is a diagonal net converging to $x$; i.e., the net $(x^i_j)_{i\in I,\,j\in J_i}$, ordered lexicographically by $I$, then by $J_i$, has a subnet which converges to $x$.
How do you prove this fact ?
Here's my progress:
Firstly I found the wording of the problem to be a bit tad too confusing -- here's what I understand:
So we have a directed set $D$, and a net $N$ on $X$, and for each $x_\lambda \in N$, there's a directed set $D_\lambda$ and $N_\lambda$ on $X$ such that it converges to $x_\lambda$. Define a new directed set $D' = \{ (i, j_i) | i \in D, j_i \in D_i \}$ such that $(\lambda_1, \mu_{\lambda_1}) \leq (\lambda_2, \mu_{\lambda_2})$ iff $\lambda_1 \leq \lambda_2$ and $\mu_{\lambda_1} \leq \mu_{\lambda_2}$ and the new net $N' = \cup_{x_\lambda \in N} N_\lambda$ such that if $i \in D$ corresponds to $X_i \in X$, and $j_i \in D_i$ corresponds to $X_{j_i} \in X$, then $(i,j_i) \in D'$ corresponds to $X_{j_i}$.
We wish to find a subnet of $N'$ such that it converge to $x$ (the original net $N$ converges to $x$)
Let $\mathfrak{U}$ be the collection of open sets around $x$. Define a function $f: N \mapsto P(\mathfrak{U})$ as follows: $f(x_\lambda) = \{ U \in \mathfrak{U} | x \in U, \lambda \leq \lambda' \Rightarrow x_{\lambda'} \in U \}$. On the other hand, each $U \in f(x_\lambda)$ is an open set containing $x_\lambda$ too, so by the convergence of $N_\lambda$, we can define a function $g_\lambda: f(x_\lambda) \mapsto D_\lambda$ such that if $g_\lambda(U) \leq \mu$, then $x_\mu \in U$. Now it would have been nice to define $\displaystyle h_\lambda > g_{\lambda}(U) $ for all $U \in f(x_\lambda)$, and let $D'' = (\lambda, h_\lambda)$ and take $N''$ to be correspodning subnet induced by $D''$.
But the problem here is that $h_\lambda$ may note be defined when $|f(x_\lambda)|$ is infinite, and this is where I'm stuck.
Some ideas to put you on the right track:
The directed set Willard describes from the directed sets $(I, \le_I)$ and the family of directed sets $(J_i, \le_{j_i})$, $i \in I$ (I'll forget about the subscripts on $\le$ in the sequel if they're clear from context) is rather:
$D'=\{(i, j): i \in I, j \in J_i\}$ with order
$$(i,j) \le (i',j') \iff i < i' \text{ or } (i=i' \land j \le j')$$
so lexicographically, where the $i$ decides the order first (if $i \neq i'$), and next (if they're equal) the order in the $J_i$ (to which then both $j,j'$ must belong by definition). Fact to check: $D’$ is in fact a directed set too. Quite easy to see.
If we have a net $f: I \to X$ and for each $i \in I$, nets $g_i: J_i \to X$, where $x \in \lim_i f$ and $f(i)=x_i \in \lim_j g_i$, we can define a diagonal net $h: D' \to X$ by
$$h(i,j) = g_{f(i)}(j)$$
And we need a subnet of $h$ that converges to $x$, so it suffices to show that $x$ is a cluster point of $h$ (by Willard's theorem 11.5) Can you show that? It's quite straightworward from the definitions.