Is $[0,1]^{[0,1]}$ Hausdorff and first-countable?

Question

I'm trying to determine if $[0,1]^{[0,1]}$ is Hausdorff or first-countable. What I know until now, is that $[0,1]^{[0,1]}$ has the product topology, then if $x\in [0,1]$ and $U$ open in $[0,1]$ the open sets of the product space are given by

$$S(x,U)=\{f:[0,1]\to[0,1]:f(x)\in U\}$$

So the $B_Y$ is a basis for $[0,1]$ the subbasis will be given by

$$\{S(x,B):x\in[0,1], B\in B_y\}$$

After that I was unable to conclude any of the questions I wanted to know. A few things are un my mind about it: Since there is $[0,1]$ is first countable seems reasonable that $[0,1]^{[0,1]}$ be first countable, I tried taking a countable basis for both but it didn't seem to work.

And to prove that is a Hausdorff space, the idea was the same that to prove $[0,1]^{[0,1]}$ is first countable, this is, using that both sets are Hausdorff. Considering $f_1,f_2\in [0,1]^{[0,1]}$ then there are $U_1,V_1\in [0,1]$ (domain) such that satisfies the Hausdorff condition and $U_2,V_2\in [0,1]$ codomain as well, then $U_1\times V_1$ and $U_2\times V_2$ should satisfy the Hausdorff condition as well?, would that prove it?.

Answer 1

Hausdorffness

Any product of Hausdorff spaces is Hausdorff.

In this specific case, note that if $\mathbf{x} = \langle x_\alpha \rangle_{\alpha \in [0,1]}$ and $\mathbf{y} = \langle y_\alpha \rangle_{\alpha \in [0,1]}$ are distinct, then $x_{\alpha_0} \neq y_{\alpha_0}$ for some $\alpha_0 \in [0,1]$. As $[0,1]$ is Hausdorff, there are disjoint open sets $U_{\alpha_0}, V_{\alpha_0} \subseteq [0,1]$ such that $x_{\alpha_0} \in U_{\alpha_0}$ and $y_{\alpha_0} \in V_{\alpha_0}$. If we let $U_\alpha = [0,1] = V_\alpha$ for all $\alpha \neq \alpha_0$, then $U = \prod_{\alpha \in [0,1]} U_\alpha$ and $V = \prod_{\alpha \in [0,1]} V_\alpha$ are open subsets of $[0,1]^{[0,1]}$, $\mathbf{x} \in U$, $\mathbf{y} \in V$, and since their $\alpha_0$-sections are disjoint, $U \cap V = \varnothing$.

It should be clear how to generalise this for an arbitrary product of Hausdorff spaces.

First countability

The product of uncountably many "non-trivial" spaces cannot be first-countable.

In this specific case, fix any $\mathbf{x} = \langle x_\alpha \rangle_{\alpha \in [0,1]} \in [0,1]^{[0,1]}$ and let $\{ U_n : n \in \mathbb{N} \}$ be any family of open neighbourhoods of $\mathbf{x}$. As each open neighbourhood of $\mathbf{x}$ includes a basic open neighbourhood of $\mathbf{x}$, we may assume without loss of generality that each $U_n$ is of the form $\prod_{\alpha \in [0,1]} U_{n,\alpha}$ where each $U_{n,\alpha}$ is open in $[0,1]$, and for each $n$ the set $\{ \alpha \in [0,1] : U_{n,\alpha} \neq [0,1] \}$ is finite.

Note that the set $\bigcup_{n \in \mathbb{N}} \{ \alpha \in [0,1] : U_{n,\alpha} \neq [0,1] \} = \{ \alpha \in [0,1] : (\exists n \in \mathbb{N} ) ( U_{n,\alpha} \neq [0,1] ) \}$ is countable. As $[0,1]$ is uncountable, there must be an $\alpha_0 \in [0,1]$ such that $U_{n,\alpha_0} = [0,1]$ for all $n \in \mathbb{N}$. Set $V_{\alpha_0} = ( x_{\alpha_0} - 1 , x_{\alpha_0} + 1 )$ (or any other open neighbourhood of $x_{\alpha_0}$ that is not all of $[0,1]$), and for $\alpha \neq \alpha_0$ set $V_\alpha = [0,1]$. Then $V = \prod_{\alpha \in [0,1]} V_\alpha$ is a neighbourhood of $\mathbf{x}$. Note that for each $n \in \mathbb{N}$, as $U_{n,\alpha_0} \nsubseteq V_{\alpha_0}$ (and each $U_{n,\alpha}$ is nonempty) it must be that $U_n \nsubseteq V$. Therefore $[0,1]^{[0,1]}$ is not not first-countable at $\mathbf{x}$ (or anywhere, as $\mathbf{x}$ was arbitrarily chosen).

Again it should be clear how to proceed in the general case.