How to prove that if a sequence converges, then a "corresponding" family of sets converges?

Question

Definition:

(convergence for a family of sets)

$A \to x \iff \forall U_x\in U(x), \exists A' \in A: A' \subset U_x$, Here $U(x)$ is a Neighbourhood Filter of $x$.

Given that:

1. Metric space $(R,d)$ is given.
2. $\{A'_i\}$ is a sequence of non-empty sets on $R$, belonging to $A$.
3. $\forall A'_i, A'_{i+1} \in A: A'_{i+1} \subset A'_i$ (they are stacked).
4. $\forall A'_i: \operatorname{diam}(A'_i) \lt \frac{1}{i}$ (here $\operatorname{diam}(A'_i) := \sup\{d(x,y)\} \mid x,y \in A'_i \} $).
5. There exists a sequence $\{a_i\}$, where every $a_i$ is taken as an arbitrary point from a corresponding $A'_i$.
6. $\{a_i\}$ converges to some point $x$ in $X$.

Prove that:

$A \to x$ holds.

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Attempt: (unsure if it's right)

For an arbitrary $\varepsilon_1 > 0$, find $m$, such that:

$$d(x, a_m) + \operatorname{diam}(A_m) < \varepsilon_1$$

Thus we can set some $\varepsilon_2 > 0$, where $\varepsilon_1 > \varepsilon_2 > 0$ to get:

$$d(x, a_m) + \operatorname{diam}(A_m) < \varepsilon_2 + \frac{1}{m} < \varepsilon_1$$

This leads us to an $m \in N$ value for any given $\varepsilon_1 > \varepsilon_2 > 0$:

$$m > \frac 1 {\varepsilon_1 - \varepsilon_2}\tag 1$$

So, for any such $m$ (wrt. to $(1)$) we would be able to find such $A'_m$ that the distance from $x$ to $a_m$ plus the distance from $a_m$ to the farthest boundary of $A'_m$ would be smaller than $\varepsilon_1$. Therefore, for every neighbourhood of $x$ there would exist $A'_m \in A$ fully contained in this neighborhood, which would satisfy the definition $A \to x$. $\square$

Answer 1

Your proof is basically correct, but I think it can be refined:

Let $\epsilon > 0$. We want to fix $A' \in A$ with $A' \subset B(x, \epsilon)$. Since $a_i \to x$, we can fix $M \in \mathbb{N}$ such that $d(x, a_i) < \epsilon/2$ for all $i \geq M$. Similarly, we can fix $N \in \mathbb{N}$ such that $\mathrm{diam}(A_i') < \epsilon/2$ for all $i \geq N$. Take $i \geq \max(M, N)$. To see that $A_i' \subset B(x,\epsilon)$, let $a \in A_i'$ be arbitrary. Then by the triangle inequality,

$$ d(x,a) \leq d(x, a_i) + d(a_i, a) \leq d(x, a_i) + \mathrm{diam}(A_i') < \epsilon, $$

as needed.