Does there exist a compact topological space not having countable basis?

I have constructed a product space from uncountably many unit intervals $[0,1]$, endowed with the product topology. Tychonoff's Theorem shows that this topological space is compact Hausdorff space, but I'm not sure how to prove that this space does not have any countable basis. Need help!

Any other suggestions on constructing such topological space (using Tychonoff's theorem) would be a huge help, too!

  • 3
    $\begingroup$ Well compact and second countable imply sequentially compact and your space is not sequentially compact. $\endgroup$ – Seth Apr 10 '14 at 23:12

A direct argument showing that $[0,1]^I$ is not second-countable for uncountable $I$ is not too difficult to drum up.

First consider the standard basis $\mathcal{B}$ for the product topology: all products of the form $\prod_{i \in I} U_i$ where each $U_i$ is open in $[0,1]$ and $U_i = [0,1]$ for all but finitely many $i \in I$.

A nice result is that given any basis of a topological space, you can always find a subset of that basis which itself is a basis, and of minimum possible size. So, if $[0,1]^I$ were second-countable, there would be a countable subset of $\mathcal{B}$ which is also a basis.

Letting $\mathcal{B}^\prime = \{ U^{(k)} : k \in \mathbb{N} \}$ be any countable subset of $\mathcal{B}$, for each $k \in \mathbb{N}$ we can write $U^{(k)} = \prod_{i \in i} U^{(k)}_i$ as above. For each $k$ let $I_k = \{ i \in I : U^{(k)}_i \neq [0,1] \}$. Each $I_k$ is finite, and so $\bigcup_{k \in \mathbb{N}} I_k$ is countable. Thus there is some $i_* \in I \setminus \bigcup_{k \in \mathbb{N}} I_k$. Letting $V_{i_*} = [0, \frac{1}{2} )$, and $V_i = [0,1]$ for all $i \neq i_*$, consider the open set $V = \prod_{i \in I} V_i$. It is not difficult to show that $U^{(k)} \not\subseteq V$ for all $k \in \mathbb{N}$, and so $\mathcal{B}^\prime$ cannot be a basis!

(By making the appropriate changes to this proof, you can show that the product of uncountably many spaces that don't have the trivial topology is not second-countable.)

For another basic example of a compact space which is not second-countable, consider the ordinal space $\omega_1+1 = [0, \omega_1]$ (where $\omega_1$ is the least uncountable ordinal). With a little knowledge of ordinals it is not too difficult to show that this space is not first-countable (there is no countable base at $\omega_1$), and therefore is not second countable.


Here's a very easy way to check that certain spaces cannot have a countable basis (though it will not detect all spaces that lack a countable basis). Note that if $\mathcal{B}$ is a basis for a topological space $X$, then for every open set $U\subseteq X$ there exists a subset $\mathcal{A}\subseteq\mathcal{B}$ such that $U$ is the union of the elements of $\mathcal{A}$. This means that there can be at most $2^{|\mathcal{B}|}$ different open subsets of $X$. So if $X$ has a countable basis, it can have at most $2^{\aleph_0}$ open subsets.

This makes it easy to find compact spaces with no countable basis. For instance, let $X$ be any compact $T_1$ space with more than $2^{\aleph_0}$ points (e.g., a product of copies of $[0,1]$, as long as you have enough copies that the cardinality of the product is greater than $2^{\aleph_0}$). Then $X$ cannot have a countable basis, since for each $x\in X$ there is a distinct open set $X\setminus\{x\}$.


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