# Dealing with Tychonoff's Theorem.

Here are my few questions that I encountered while going through Tychonoff's theorem in Wikipedia.

a) First of all, so far I was thinking that Heine Borel definition of compactness implies sequential compactness but not the other way around ( although i am failing to find some examples to appreciate it). But what wikipedia says is that " but NEITHER implies the other in general topological space . What am i missing here ?

b) It is easy to see that finite product ( countable product is not true, right ? ) of sequentially compact spaces is compact which we can see using diagonalization argument . and it discusses of embedding X ( completely regular Hausdorff space ) into $[0,1]^{C(X,[0,1])}$ (what does $[0,1]^{C(X,[0,1])}$ mean? I am not able to make any sense) , where $C(X,[0,1])$ is the set of continuous map from $X$ to $[0,1]$. I would appreciate your help.
Thanks!

• $A^B$ means $\{ f : B \to A \}$. – Patrick Da Silva May 16 '12 at 16:43
• Thanks a lot. But what would a map from $[0,1]$ to the set of continuous functions imply ? Can you explain a bit what is happening in the example ? – Theorem May 16 '12 at 16:48
• It would mean that for instance, $f : [0,1] \to C([0,1])$ maps some $i \in [0,1]$ to a continuous function $f_i(t)$. Saying that $f$ is continuous means that $f_i(t)$ varies continuously as $i$ varies. Of course, this notion of continuity depends on the topology you put on $C([0,1])$, but that's another story. – Patrick Da Silva May 16 '12 at 17:16
• What do you mean by Heine-Borel definition of compactness? – T. Eskin May 16 '12 at 17:24
• @ThomasE. It's the usual definition of compactness, using open covers. – M Turgeon May 16 '12 at 18:12

$a):$ What wikipedia says is correct, open-cover compactness does not imply sequential compactness and vice versa in general topological spaces: in metric spaces they are equivalent.

For example, denote $X=\{0,1\}^{\mathbb{N}}$, which is open-cover compact by Thychonoff's theorem, and moreover $Y:=\{0,1\}^{X}$ is open-cover compact by the same argument. But $Y$ is not sequentially compact. The projections $pr_{n}:X\to \{0,1\}$ are members of $Y$ and the sequence $(pr_{n})_{n=1}^{\infty}$does not have a converging subsequence. Assume the contrary that some subsequence $(pr_{n_{k}})_{k=1}^{\infty}$ has a limit. This means, that for each $x\in X$ the sequence $(x_{n_{k}})_{k=1}^{\infty}$ converges in $\{0,1\}$. Let $x\in X$ be a sequence such that $x_{n_{k}}=1$ when $k$ is even and $x_{i}=0$ when $i=n_{k}$ for some odd $k$ and when $i$ is not of the form $n_{k}$. Then it follows that a sequence of the form $0,1,0,1,...$ converges which is a contradiction. Hence $(pr_{n})_{n=1}^{\infty}$ has no converging subsequences.

As an example of a sequentially compact space which is not cover-compact you may consider $\omega_{1}$ with the order topology. It is not even Lindelöf. This example is quite standard I believe, I would fill in more details if it wasn't so late here where I live. (I might edit some details later.)

$b):$ I think this question was answered in the comments below your question.

The main thing to understand (and you will see this comment over and over again) is that general topology is wild, and that statements that seem intuitive will sometimes be false, and counterexamples will be given by highly "non-regular" spaces. For a convenient source of counterexamples, I recommend Counterexamples in Topology by Steen & Seebach.

1. Compactness is not equivalent to Sequential Compactness, and vice-versa. For a compact space which is not sequentially compact, the main example is $[0,1]^{[0,1]}$ (with the product topology). On the same page, you will find an example of a sequentially compact space that is not compact. What this is highlighting, is that sequences are not the right thing to study in general, but one should rather study nets.
2. First of all, a countable product of sequentially compact spaces is sequentially compact (you can find a proof here). As for the last part of your question, and as Patrick noted, the notation $[0,1]^{C(X;[0,1])}$ means all the functions from the space of continuous functions $f:X\to [0,1]$ to the interval $[0,1]$. The result mentioned is related to the Stone-Cech compactification theorem.