Prove or disprove that a certain net exists I am trying to prove or disprove the following claim:
There exists a net $(m_{\alpha})_{\alpha \in A}$ in $\mathbb{N}$, such that $cos(2\pi m_{\alpha}x) \rightarrow 1$ for every $x \in \mathbb{R}$.
I have tried one strategy and found that there exists a net $(m_{\alpha})_{\alpha \in A}$ in $\mathbb{N}$ such that $cos(2\pi m_{\alpha}x) \rightarrow f(x)$ for every $x \in \mathbb{R}$, where $f$ is an element of the product space $\Pi_{x \in \mathbb{R}} [-1, 1]$, but so far I haven't been able to show the result that I want.
Any ideas or hints would be appreciated!
 A: I'm not sure if the claim is true, but here is an approach which seems to almost make it (but doesn't, as written). Unfortunately there is some conflict of notation with $\{\}$ indicating both sets and fractional part, but I hope it is not confusing.
Let the index set be $\mathcal A\times\mathbb N$ where $\mathcal A$ is the collection of all finite subsets of $[0,1]$. The ordering is $(A_1, k_1)\leq (A_2, k_2) \iff A_1\subseteq A_2$, and $k_1\leq k_2$. You can check that this is a directed set.
We'll use that for any irrational number $x$, $\{\{\ell x\}\mid \ell\in\mathbb N\}$ is dense in $[0,1]$. 
I make the assumption now that even with a finite number of irrational numbers, say $\alpha_1,\ldots,\alpha_m$, we can find $\ell$ such that $\max\{\{\ell\alpha_1\}, \ldots,\{\ell\alpha_m\}\} < \epsilon$ for a given $\epsilon > 0$.
(Kronecker's theorem says this is true if the $\alpha_i$ are linearly independent over $\mathbb Q$, but I do not know what happens otherwise.)
Hence given $A=\{x_1,\ldots, x_m\}$, we can find a sequence $\ell_{1,A}, \ell_{2,A}, \ldots$ such that $\{\ell_{k,A}x\}< \frac1k$ for every $k$ and $\forall x\in A$. Also note that this sequence depends only on $\{x_i\}$'s (fractional parts) and not on $x_i$'s itself. 
Now we define our net: $\alpha:=(A, k) \mapsto \ell_{k,A} := m_\alpha$.
It's simple to check now that $\cos(2\pi m_{\alpha} x)\to 1$ for every $x$. Given the $x$ and an $\epsilon>0$, let $\delta$ be such that $\{x\}<\delta \implies \cos(2\pi\{x\}) >1- \epsilon$.
Take $\frac1k<\delta$, and consider $\alpha_0 = (\{x\}, k)$. Then if $\alpha = (\{x,x_1,\ldots, x_n\}, k') \geq \alpha_0$, clearly $\{m_\alpha x\} < \delta$, so $\cos(2\pi m_\alpha x) >1- \epsilon$.
