Uncountable Cartesian product of closed interval 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?

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.
