# Product space that is compact, but isn't sequentially compact

I need to find an example of infinite product space that is compact, but it's not sequentially compact.

So this is my example:

$$\prod_{i\in [0,1[}\{0,\ldots,9 \}.$$

It sure is compact, but is it sequentially compact?

Let's define a sequence $(f_{n})_{n\geq 1}$ by

$$f_{n}:[0,1[\rightarrow \{0,\ldots ,9\},\ f_{n}(x)=a_{n},$$ where $x\in [0,1[$ has decimal expansion $0.a_{1}a_{2}a_{3}a_{4}...$.

Are there more simple examples than this?

• It's the decimal sequence of number $x$, $0.a_{1}a_{2}a_{3}...$
– Zzz
Dec 3 '15 at 18:01
• I doubt there is any example that is significantly simpler than this one, assuming you want the individual factors in the product to be sequentially compact. Also, I don't know what $x=0.999...$ has to do with anything. Dec 3 '15 at 18:08
• I think I got it all wrong. I thought that the 0.999... represents the sequence that doesn't have converging subsequence. Infinite product spaces are new area for me.
– Zzz
Dec 3 '15 at 19:56
• If you're familiar with the Stone-Cech compactification, then $\beta\mathbb N$ (the Stone-Cech compactification of the natural numbers) is a compact space and the sequence $x_n=n$ has no convergent subsequences.
– SamM
Dec 3 '15 at 20:50
• No, the counter example is the sequence $(f_n)$, which has no convergent subsequence. A sequence in the product converges iff it converges pointwise Dec 3 '15 at 20:51

An example which is homeomorphic to yours, but does not depend on expansion of reals etc: take the index set $I = \{0,1\}^{\mathbb{N}}$, the set of all $0$-$1$-sequences (the Cantor set, to a topologist). This has natural functions $\pi_n$ that sends an $i \in I$ (which is really a sequence, i.e itself a function from $\mathbb{N}$ to $\{0,1\}$) to its $n$-th coordinate; this corresponds to the $n$-th decimal of the real number $i$ in your example.

Now consider $X = \{0,1\}^I$. This is compact by Tychonoff's theorem. Define $f_n \in X$ (for every (fixed for now) $n$) as $f_n(i) = \pi_n(i)$. This is well-defined, because a member of $X$ is just a function from $I$ to $\{0,1\}$, and for some fixed $n$ we have that we send $i \in I$ to its $n$-th coordinate, which is $0$ or $1$. So we have a sequence $(f_n)$ in $X$.

A basic fact for the product topology, for any sequence $(x_n)$ in $X$ (so all $x_n$ are functions from $I$ to $\{0,1\}$!): $(x_n) \rightarrow x (\in X)$ iff for all $i \in I$, $x_n(i) \rightarrow x(i)$. So convergence is determined pointwise. In fact we only need the implication from left to right (which just says that all projections are sequentially continuous).

Suppose that $(f_n)_n$ has a convergent subsequence $f_{n_k}$. Define a point $j \in I$ by $j_{n_{2k}} = 1$ for all $k$, $j_m = 0$ for all other $m$.

Then $f_{n_k}(j) = \pi_{n_k}(j) = j_{n_k}$ which is just (as $k$ goes from $0,1,\ldots$) the sequence $1, 0, 1, 0, 1, 0, \ldots$ by how $j$ was defined. So this subsequence does not converge pointwise at $j$ so cannot converge in $X$ at all. Contradiction.

So $X$ is not sequentially compact.

It turns out that if $X_i$ are non-trivial (more than 1 point) sequentially compact Tychonoff spaces for $i \in I$, then $\prod_i X_i$ is sequentially compact for countable index set $I$, and never sequentially compact for $I$ of size $\mathfrak{c}$. There is some turning point cardinal in between, related to the cardinal invariant called the splitting number. If CH holds, of course all is known, but in general we could have that e.g. $\{0,1\}^{\aleph_2}$ is sequentially compact, while $\{0,1\}^{\aleph_3}$ is not, even with $\aleph_3 < \mathfrak{c}$, etc. This as a set-theory aside and it shows that we need "big" products to kill sequential compactness in products.