Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 quiet understand the proof given on the book. Can someone give a more detail proof?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

It is not.

Let's look at open sets containing 0 (sequence of 0s).

We will argue by contradicton, so suppose there is a local neighbourhood 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 sum 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 contained in any of $U_i$, contradicting first-countability at 0.

share|improve this answer
The proof is much clear to me than the one given in the book. Thanks! –  Qomo Sep 26 '11 at 5:10

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.

share|improve this answer
Thanks for your proof! Much more clear right now. –  Qomo Sep 26 '11 at 5:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.