Net Convergence Let $\Gamma$ be a uncountable set and $\psi(\Gamma)\subset\mathbb{R}^{\Gamma}$ be the space of all countably supported elements. How can I show that $\overline{\psi(\Gamma)}=\mathbb{R}^\Gamma$, i.e. for every $x\in\mathbb{R}^\Gamma$, there exist a net in $\psi(\Gamma)$ which converges to $x$ in the product topology. 
 A: Do you want to use nets absolutely? Only using neighborhoods:
Let $(x_{\gamma}) \in \mathbb{R}^{\Gamma}$ and $U$ an open neighborhood of $(x_{\gamma})$. From the definition of the product topology, we deduce that there exist open sets $O_{\gamma}$ in $\mathbb{R}$ such that $\prod\limits_{\gamma \in \Gamma} O_{\gamma} \subset U$ and $O_{\gamma}=\mathbb{R}$ for all but finitely many $\gamma \in \Gamma$. Now, for $\gamma \in \Gamma$, take $y_{\gamma}=x_{\gamma}$ if $O_{\gamma} \neq \mathbb{R}$ and $y_{\gamma}=0$ otherwise; then $(y_{\gamma}) \in \psi(\Gamma) \cap \prod\limits_{\gamma \in \Gamma} O_{\gamma} \subset \psi(\Gamma) \cap U$. So $\psi(\Gamma)$ is dense in $\mathbb{R}^{\Gamma}$; in fact, we proved that the subspace of all finitely supported elements is dense in $\mathbb{R}^{\Gamma}$.
A: Let $\mathcal D$ denote the family of all countable subsets of $\Gamma$, and note that inclusion $\subseteq$ is a direct order on $\mathcal D$.
Given any $x = \langle x_\gamma \rangle_{\gamma \in \Gamma} \in \mathbb{R}^\Gamma$, define a net $( y^A = \langle y^A_\gamma \rangle_{\gamma \in \Gamma} )_{A \in \mathcal{D}}$ as follows: $$y^A_\gamma = \begin{cases}
x_\gamma, &\text{if }\gamma \in A \\
0, &\text{otherwise.}
\end{cases}$$  It should be easy to show that this net converges to $x$, and trivial to see that each $y^A$ belongs to $\psi ( \Gamma )$.
