I have a question about product topology.
Suppose $I=[0,1]$, i.e. a closed interval with usual topology. We can construct a product space $X=I^I$, i.e. uncountable Cartesian product of closed interval. Is $X$ first countable?
I have read Counterexamples of Topology, on item 105, it is dealing with $I^I$. I do not quite understand the proof given on the book. Can someone give a more detail proof?
On
Let $x \in I^I$ and assume $I^I$ has a countable basis $\langle A_n \rangle_{n\in \omega}$ at $x.$
Let $s_n$ be the set $i\in I$ such that the $i$-th coordinate projection map $\pi_i: A_n \rightarrow I$ is not surjective. As each $A_n$ is open, the set $s_n$ is finite for all $n.$ It follows $s := \cup_{n\in \omega} s_n$ is countable.
As $I$ is uncountable, it follows $I\setminus s$ is nonempty. Choose an element $i\in I\setminus s.$ Define for each $j\in I$ a set $X_j$ such that $X_j = I$ if $j \neq i$ and $(\pi_i(x) - 1/2,\pi_i(x) + 1/2) \cap I$ otherwise.
Then $X := \prod_{j \in I} X_j$ is open and contains $x$ but is not contained in $A_n$ for any $n.$ This contradicts the fact $\langle A_n \rangle_{n\in \omega}$ was assumed to be a basis at $x.$
It follows $I^I$ is not only not first countable but $I^I$ doesn't have a countable basis at any point.
It is not.
Let's look at open sets containing 0 (sequence of 0s).
We will argue by contradiction, so suppose there is a countable local neighborhood basis $U_i$ for 0. Every such $U_i$ contains some $V_i = \prod_{r} V_{i,r}$ where $V_{i,r} = I$ for almost every $r$, $V_{i,r}$ open. (By definition of product topology).
Let's look at the set of all $r$ that $V_{i,r} \neq I$ for some $i$.
This set is countable, because it's a countable union of countable sets. So it's not the whole of $I$. Let's choose some $r_0$ outside this set.
Let $H = \prod_r H_r$ where $H_{r_0} = [0, 1/2)$ and $H_{r} = I$ otherwise.
Then $H$ is an open set containing $0$, not containing any of $U_i$, contradicting first-countability at 0.