Locally Compact Spaces Consider $\mathbb{R}^{\omega} = \{(x_n): x_n \in \mathbb{R} \ \forall n \in \mathbb{N} \}$ with the product topology, induced by the standard topology on $\mathbb{R}$. Show that $\mathbb{R}^{\omega}$ is not locally compact.
So far I've proven the following: $\overline{\prod_n{A_n}} = {\prod_n{\overline{A_n}}}$ by the following process:
Let $(x_n)_{n \in \mathbb{N}}$ be a closure point of $\prod_n{A_n}$.
Consider, for $\beta \in \mathbb{N}$, a neighborhood $U_{\beta}$ of $x_\beta$. Since the projection maps are continuous, $\pi^{-1}_{\beta} (U_{\beta})$ is open in $\prod_n{X_n}$, hence a neighborhood of $(x_n)_{n \in \mathbb{N}}$ so it contains a point $(y_n)_{n \in \mathbb{N}} \in \prod_n{A_n}$. 
In particular $y_{\beta} \in U_{\beta} \cap A_{\beta}$.
So $x_{\beta} \in \overline{A}_{\beta}$
Conversely, let $(x_n)_{n \in \mathbb{N}} \in \prod_n{\overline{A}_n}$ and $U = \prod_n{U_n}$, a neighborhood of $(x_n)_{n \in \mathbb{N}}$
Then, every $U_n$ contains a point $y_n \in U_n \cap A_n$, so $(y_n)_{n \in \mathbb{N}} \in U \cap \prod{A_n}$, which means $(x_n)_{n \in \mathbb{N}} \in \overline{\prod_n{{A}_n}}$
Can someone help with where I should go next?
 A: I can't tell you where to go next because I don't know where you are now. I mean, I have no idea what those steps you wrote have to do with your goal of showing that $\mathbb R^\omega$ is not locally compact. Let's back up to the starting point. I will sketch a proof.
We can show that $\mathbb R^\omega$ is not locally compact by showing that some point $a\in\mathbb R^\omega$ does not have a compact neighborhood. It shouldn't matter what point we use for $a,$ the points are all the same; for concreteness, take the point whose coordinates are all zero.
Assume for a contradiction that $a$ has a compact neighborhood, call if $K.$ Then $K$ contains a basic neighborhood of $a,$ call it $B.$ According to the definition of the standard basis for the product topology, $B$ depends on only a finite set $I$ of coordinates. Since $\mathbb N$ is infinite, we can choose an index $i\in\mathbb N\setminus I.$
Consider the projection function $\pi_i:\mathbb R^\omega\to\mathbb R$ which maps each point $x\in\mathbb R ^\omega$ to its coordinate $x_i.$ Since $\pi_i$ is continuous and $K$ is compact, the image set $\pi_i(K)$ is a compact subset of $\mathbb R.$ Now get a contradiction by observing that $\pi_i(K)\supseteq\pi_i(B)=\mathbb R.$
