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In a ps file on topological properties, the set $[0,1]^{[0,1]}$ is given as an example of a product topology that is not first countable.

Is there a proof of why?

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3 Answers 3

You can also proceed indirectly, by showing that there exist a real-valued function on some subspace of $[0, 1]^{[0, 1]}$ that is sequentially continuous but not continuous. It is the case of

$$F(f)=\int_0^1 f(t)\, dt,\qquad f\ \text{measurable}.$$

See this answer by Auguste Hoang Duc. (You can also tweak this construction a little to find some set for which the sequential closure does not match with the topological closure. I think that this would yield something similar to Brian's answer above).

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More generally, if $X$ is any space that has a subspace homeomorphic to the two-point discrete space, and $J$ is any uncountable set, then $X^J$ is not first countable.

Suppose that $X$ and $J$ are as described, and let $D$ be a two-point subset of $X$ on which the relative topology is discrete. (In the case of $X = [0,1]$, you could for instance let $D = \{0,1\}$, or indeed any other two-point subset of $[0,1]$.) First countability is a hereditary property, so if $X^J$ were first countable, $D^J$ would also be first countable; we’ll show that it isn’t.

Let $D = \{p,q\}$, and let $x \in D^J$ be defined by $x(j)=p$ for all $j\in J$. For each finite $F \subseteq J$ let $B(F) = \{y \in D^J:y(j) = p\text{ for every }j \in F\}$, and let $\mathscr{B}$ be the collection of all of the $B(F)$; by definition $\mathscr{B}$ is a local base at $p$ in $D^J$.

For each finite $F \subseteq J$ let $y_F \in D^J$ be the point defined by $$y_F(j) = \begin{cases}p,&\text{if }j\in F\\q,&\text{otherwise},\end{cases}$$ and let $A = \{y_F:F \subseteq J\text{ is finite}\}$. Clearly $x \notin A$. On the other hand, it’s clear that $y_F \in B(F)$ for each $B(F) \in \mathscr{B}$, so $x \in \operatorname{cl}A$.

If $D^J$ were first countable, this would imply that $x$ was the limit of a sequence of points of $A$. Suppose, to get a contradiction, that $\langle y_{F(n)}:n \in \mathbb{N} \rangle$ is a sequence in $A$ converging to $x$; then every open nbhd of $x$ contains all but finitely many terms of the sequence. Let $I = \bigcup\limits_{n \in \mathbb{N}}F(n)$; each $F(n)$ is finite, so $I$ is a countable subset of $J$, and we may choose an index $j \in J \setminus I$. Then $y_{F(n)}(j) = q$ for every $n \in \mathbb{N}$, so $B(\{j\})$ is an open nbhd of $x$ that contains no point of the sequence, and we have our contradiction. $\dashv$

We didn’t actually need the relative topology on $D$ to be discrete: we just needed to be sure that $\{p\}$ was open in $D$. (Of course if $X$ is a $T_1$-space, $D$ will automatically inherit the discrete topology.) We also didn’t need all of the factor spaces to be identical. Essentially the same proof shows that if $\{X_j:j \in J\}$ is any uncountable collection of spaces, and if uncountably many of these spaces contain a subspace $D_j = \{p_j,q_j\}$ in which $\{p_j\}$ is a relatively open set, then $\prod\limits_{j\in J} X_j$ is not first countable.

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I'm assuming that by $[0,1]^{[0,1]}$ you mean a product of $[0,1]$ of cardinality of the continuum, which I will take as being homeomorphic to $[-1,1]^{[0,1]}$ for convenience. Take a local basis around the origin. Each of the sets in the local basis contains all but finitely many of the axes, so no countable collection of open sets can avoid having all of them contain one of the axes, say $\alpha$. But we know that $\Pi_\beta U_\beta$, where $U_\beta = [-1,1]$ when $\beta \ne \alpha$ and $U_\alpha=(-1/2,1/2)$ is a neighborhood of the origin. Hence any local basis must be uncountable.

I've left a couple things for you to fill in, but these are just using standard tricks to switch between the standard basis for the product topology and an arbitrary basis.

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It might help to clarify what you mean by "axes" in this context. –  Nate Eldredge Sep 18 '11 at 13:07
    
True. By axes, I mean the sets of the form $\Pi_\beta K_\beta$, where $K_\beta = \{0\}$ for all $\beta \in [0,1] - \{c\}$, and $K_c = [-1,1]$, for some $c \in [0,1]$. –  Logan Maingi Sep 18 '11 at 17:25

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