Deduce that the product of uncountably many copies of the real line $\mathbb{R}$ is not metrizable.
Let $J$ be an uncountable set. Suppose that for $x = (x_j)_{j \in J} \in \prod_{j \in J} \Bbb{R}$ there exist $\{U_n\}_{n \in \Bbb{N}}$ satisfying the conditions for first countability.
By the definition of product topology, for each $U_i = \prod_{j \in J} U_{i,j}$, there are at most finitely many $U_{i,j} \not= \Bbb{R}$. Let $k_i = \{ j \in J | U_{i,j} \not= \Bbb{R} \}$. So $k_i$ is a finite set. Now let $K = k_1 \cup k_2 \cup …$. Since $K$ is the countable union of countable sets, it must be countable.
Define $f: K \rightarrow J$ to be the inclusion. Since $K$ is countable and $J$ is not, $f$ cannot be surjectie. i.e. there exists $l \in J$ such that $f^{-1}(l)$ is the empty set. If we pick a neighborhood $V = \prod_{j \in J} V_j$ with $V_l \subsetneq \Bbb{R}$, then for all $i \in \Bbb{N}$, $U_i \nsubseteq V$. A contradiction.
Is my proof correct?
Best Answer
It looks perfect to me, and well written, too.