general topology - Net convergence vs sequence convergence

I am stuck trying to solve the following exercise regarding nets and sequences:

Let $(x_n)_{n\in \mathbb N}$ a sequence in the metric vector space $(V,d)$. Let $\mathcal M$ be the set containing all finite subsets of $\mathbb N$ and define the relation $\leq$ on $\mathcal M$ by $$\forall M, N \in \mathcal M: \quad M\leq N:\Leftrightarrow M \subseteq N.$$
For $N \in \mathcal M$ and $k \in \mathbb N$ define $$s_N := \sum_{n\in N} x_n \quad\text{ and }\quad \tilde s_{k} := \sum_{i = 1}^k x_i$$
(a) Show that $s_N$ is a net in V.
(b) Show that $\lim \limits_{N\in \mathcal M} s_N = s \implies {\lim \limits_{k \to \infty}} \tilde s_k = s$
(c) Does the reverse implication hold in (b)?

My ideas:
(a) It is easy to see that $\leq$ makes $\mathcal M$ into a directed set, hence $s_N$ is a net.

(c) The reverse implication does not hold. Let, for example, $\tilde s_k$ be the alternating harmonic series on $\mathbb R$. It converges to $\ln2$. However, since we look at finite subsets of $\mathbb N$, for any given $N \in \mathcal M$ we can always choose a subset of $\mathbb N$ only containing, for example, uneven numbers, so $s_N$ does not converge. Is that correct?

(b) I think I have the right idea, but it is hard to formalize it and write it down. The sequence takes $k \to \infty$, but for the net, we can't go towards infinity since we only look at finite subsets. Please help me out.

Answer

For (b), suppose that $\lim\limits_{N\in\mathscr{M}}s_N=s$. Let $\epsilon>0$ be arbitrary; there is an $M_\epsilon\in\Bbb N$ such that $|s_n-s|<\epsilon$ whenever $N\in\mathscr{M}$ and $M_\epsilon\le N$. Let $m_\epsilon=\max M_\epsilon$; if $n\ge m_\epsilon$, then $M_\epsilon\le\{1,\ldots,n\}\in\mathscr{M}$, so $|\tilde s_n-s|<\epsilon$. Thus, $\lim\limits_{k\to\infty}\tilde s_k=s$.

Your argument for (c) is on the right track but needs a little work. One way to make it precise is to show that for any $M\in\mathscr{M}$ and any real number $a$ there is an $N_a\in\mathscr{M}$ such that $M\le N_a$ and $s_{N_a}>a$. You can do this by letting $m=\max M$ and observing that since the harmonic series diverges, there is an $\ell\in\Bbb N$ such that

$$\sum_{k=m}^\ell\frac1{2k}>a-s_{M}\;,$$

so that if we set $N_a=M\cup\left\{\frac1{2k}:m\le k\le\ell\right\}$, then

$$s_{N_a}=s_M+\sum_{k=m}^\ell\frac1{2k}>a\;.$$

Thus, the net does not converge to $\ln 2$ (or any other real number).