Example 2, Sec. 29, in Munkres' TOPOLOGY, 2nd ed: Local compactness of $\mathbb{R}^\omega$ In Example 2, Sec. 29, in the book Topology by James R. Munkres, 2nd edition, the author shows that $\mathbb{R}^\omega$ (in the product topology) is not locally compact (at any point). However, the argument he gives is not as detailed as I would like it to be. Accordingly, I would like to present the following proof:

By definition
$$
\mathbb{R}^\omega := \mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \cdots, \tag{Definition 0} 
$$
that is, $\mathbb{R}^\omega$ is the set of all the (infinite) sequences of real numbers.


Here $\mathbb{R}^\omega$ has the product topology determined by the standard (or usual) topology on $\mathbb{R}$.


Let $\mathbf{x} := \left( x_1, x_2, x_3, \ldots \right)$ be any point of $\mathbb{R}^\omega$. [Refer to (Definition 0) above.] We show that $\mathbb{R}^\omega$ is not locally compact at $\mathbf{x}$.


Suppose, if possible, that $\mathbb{R}^\omega$ is locally compact at $\mathbf{x}$. Then by definition there exists a compact subspace $\mathbf{C}$ of $\mathbb{R}^\omega$ containing a neighborhood $\mathbf{U}$ of $\mathbf{x}$, that is, $\mathbf{U}$ is an open set of $\mathbb{R}^\omega$ and $\mathbf{C}$ is a compact subspace of $\mathbb{R}^\omega$ such that
$$
\mathbf{x} \in \mathbf{U} \subset \mathbf{C}. \tag{0} 
$$


Now as $\mathbf{x} \in \mathbf{U}$ and $\mathbf{U}$ is an open set of the product space $\mathbb{R}^\omega$, so there exists a canonical basis set $\mathbf{B} := U_1 \times U_2 \times U_3 \times \cdots$ for the product topology on $\mathbb{R}^\omega$, where $U_{n_1}, \ldots, U_{n_k}$ are open sets of $\mathbb{R}$ for some finitely many natural numbers $n_1, \ldots, n_k$ and $U_n = \mathbb{R}$ for all the other natural numbers $n$, such that
$$
\mathbf{x} \in \mathbf{B} \subset \mathbf{U}. \tag{1} 
$$


Now as $\mathbf{x} = \left( x_1, x_2, x_3, \ldots \right) \in \mathbf{B} = U_1 \times U_2 \times U_3 \times \cdots$, so for each $j = 1, \ldots, k$, we have $x_{n_j} \in U_{n_j}$, and since $U_{n_j}$ is an open set of $\mathbb{R}$, there exists an open interval $\left( a_{n_j}, b_{n_j} \right)$ such that
$$
x_{n_j} \in \left( a_{n_j}, b_{n_j} \right) \subset U_{n_j}. \tag{2} 
$$
Let us now put
$$
\mathbf{B}^* := B_1 \times B_2 \times B_3 \times \cdots, \tag{Definition 1}
$$
where
$$
B_n := \begin{cases} \left( a_{n_j}, b_{n_j} \right) & \mbox{ if } n = n_j \mbox{ for some } j = 1, \ldots, k; \\ U_n = \mathbb{R} & \mbox{ otherwise}.  \end{cases} \tag{Definition 2} 
$$
Then by (Definition 2), (Definition 1), (2), (1), and (0) above, we have
$$
\mathbf{x} \in \mathbf{B}^* \subset \mathbf{B} \subset \mathbf{U} \subset \mathbf{C}, 
$$
which implies that
$$
\mathbf{x} \in \mathbf{B}^* \subset \mathbf{C},
$$
and hence
$$
\overline{\mathbf{B}^*} \subset \overline{\mathbf{C}} = \mathbf{C},
$$
because $\mathbf{C}$, being a compact subspace of the (metrizable and hence) Hausdorff space $\mathbb{R}^\omega$, is also closed in $\mathbb{R}^\omega$. Now as $\overline{\mathbf{B}^*}$ is a closed subset of $\mathbb{R}^\omega$, so $\overline{\mathbf{B}^*} \cap \mathbf{C} = \overline{\mathbf{B}^*}$ is also closed in the subspace $\mathbf{C}$ of $\mathbb{R}^\omega$. Therefore $\overline{\mathbf{B}^*}$, being a closed subspace of the compact space $\mathbf{C}$, is also compact as a subspace of $\mathbf{C}$, and since $\mathbf{C}$ in turn is a subspace of $\mathbb{R}^\omega$, therefore we can conclude that $\overline{\mathbf{B}^*}$ is also compact as a subspace of $\mathbb{R}^\omega$.


But
$$
\overline{\mathbf{B}^*} = \overline{B_1} \times \overline{B_2} \times \overline{B_3} \times \cdots, 
$$
where
$$
\overline{B_n} = \begin{cases} \left[ a_{n_j}, b_{n_j} \right] & \mbox{ if } n = n_j \mbox{ for some} j = 1, \ldots, k; \\ \mathbb{R} & \mbox{ otherwise}. \end{cases}
$$
Let $n_0$ be any natural number distinct from $n_1, \ldots, n_k$. Then
$$
\pi_{n_0} \left( \overline{\mathbf{B}^*}  \right) = \mathbb{R}. 
$$
And, as the mapping $\pi_{n_0} \colon \mathbb{R}^\omega \longrightarrow \mathbb{R}$ is continuous and as $\overline{\mathbf{B}^*}$ is a compact subspace of $\mathbb{R}^\omega$, so $\pi_{n_0} \left( \overline{\mathbf{B}^*}  \right) = \mathbb{R}$ is also compact, which is a contradiction.


Therefore our supposition that $\mathbb{R}^\omega$ is locally compact at $\mathbf{x}$ is wrong.


Hence $\mathbb{R}^\omega$ is not locally compact at any point $\mathbf{x} \in \mathbb{R}^\omega$.

Is this proof correct and clear enough? If so, is my argument any clearer than the argument given by Munikres?
This Math Stack Exchange post of mine is also somewhat relevant.
 A: Your proof is correct. But to be honest, I do not think it is better than Munkres' proof. Perhaps Munkres is a bit short and leaves some details to the reader, and you filled in these details.
Munkres shows that none of the basic open sets having the form
$$B = (a_1,b_1) \times \ldots (a_n,b_n) \times \mathbb R \times \mathbb R \times  \ldots $$
has compact closure. Here he invokes Theoem 19.5.
Why does this suffice to prove that $\mathbb R^\omega$ is not locally compact? Given $x =(x_i) \in \mathbb R^\omega$, assume that there is an open $U \subset \mathbb R^\omega$ and a compact $C \subset \mathbb R^\omega$ such that $x \in U \subset C$. Take a basic open $V = \prod_{i=1}^\infty V_i$ (i.e. $V_i  =\mathbb R$ for almost all $i$) such that $x \in V \subset U$. Choose $n$ such that $V_i = \mathbb R$ for $i > n$. Then pick open intervals $(a_i,b_i)$ such that $x_i  \in (a_i,b_i) \subset V_i$ for $i \le n$. This gives you a basic open $B$ as considered by Munkres such that $x \in B \subset C$. But we have $\overline B \subset \overline C = C$, thus $\overline B$ must be compact since it is a closed subset of a compact set.
Here is one more alternative proof.
The above assumption leads to $x \in V \subset C$ with a basic open $V$. The projections $p_i : \mathbb R^\omega \to \mathbb R, p_i(x) = x_i$, are continuous, thus the $C_i = p_i(C)$ are compact. Clearly $V_i = p_i(V) \subset C_i$. This contradicts the fact that almost all $V_i = \mathbb R$.
