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I am a beginner in topology so please do bear with me. I read that there are compact sets that are not sequentially compact, i.e. $I^I$

A sequentially compact set is countably compact, where countable open covers have finite subcovers. ... Apparently this is different from compact where all open covers have finite subcovers..

But I could not get it why the uncountable cartesian product of the unit interval $I^I$ is compact but not sequentially compact. If it is compact, then it means there is a finite subcover in $I$.. But isn't the product uncountable and infinite?

Also, suppose that there is such a finite subcover for $I^I$, say $E$.. Then isnt $E$ a countable cover that is finite in $I^I$?

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    $\begingroup$ The notation $I^I$ is ambiguous - one usually understands it as the space of all continuous functions $f : I \to I$, but it seems that you mean the infinite product of copies of $I_\alpha = I$, $\alpha \in I$. There is a general theorem (Tychonoff's theorem) which says that any product of compact spaces is compact (see any book on topology). $\endgroup$
    – Paul Frost
    Apr 29 '18 at 8:16
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    $\begingroup$ See page 3 of andrew.cmu.edu/user/calmost/pdfs/sasms_F04.pdf for a proof that $I^I$ is not sequentially compact. $\endgroup$
    – Paul Frost
    Apr 29 '18 at 8:29
  • $\begingroup$ Thanks. I checked the paper and $I^I$ is the product of all functions from [0,1] to [0,1]. I get the Tychonoff theorem part but I cant picture how to make it compact using the finite subcover idea $\endgroup$
    – Link L
    Apr 29 '18 at 9:01
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    $\begingroup$ The proof of the Tychonoff theorem (i.e. showing that every open cover has a finite subcover) is not trivial if you consider arbitrary products. Only for finite products there is reasonably elementary proof. I recommemd you to study some book on topology. BTW $I^I$ is the set, not the product, of all (not necessarily continuous) functions $I \to I$. $\endgroup$
    – Paul Frost
    Apr 29 '18 at 10:03
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    $\begingroup$ You seem to be comparing countable compactness and compactness. There we do have the obvious implication. But sequence convergence is quite different from the cover definitions. It's stronger, so harder to achieve in general. $\endgroup$ Apr 29 '18 at 13:27
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$I^I$ (i.e. the set of all functions from $I = [0,1]$ to $[0,1]$ in the product topology) is compact, because it is a product of continuum many copies of the compact (metric) space $[0,1]$ by Tychonoff's theorem. So every open cover of $I^I$ has a finite subcover. This is by no means trivial.

In particular $I^I$ is also countably compact, and this implies that a countable subset $A$ of $I^I$ has an accumulation point $p$, which means that every neighbourhood of $p$ contains infinitely many points of $A$. But sequential compactness is about convergence of sequences (every neighbourhood of the limit contains all but finitely many points of the sequence) and this is a lot harder to achieve. $I^I$ is not first-countable, in fact any local base of any point is size continuum, so it's far from a metric space in that respect, which are first countable and where we thus can describe the topology completely using sequences. In $I^I$ there are sequences such that no subsequence of it can converge at all. I describe this in my answer here (quite technical) in a so-called diagonal argument: essentially a sequence converges iff it converges coordinatewise, and we use 1 coordinate to "kill" a possible subsequence of a sequence (any sequence has exactly continuum many subsequences).

There are compact spaces that simply have no convergent subsequences except those that are eventually constant. The space $\omega^\ast$, the remainder of $\omega$ in $\beta \omega$, is such an example. In general spaces we can have compact spaces that are not sequentially compact (mostly because sequences are not enough in general spaces) and sequentially compact spaces that are not compact (mostly because sequences are inherently countable, so often we can reduce countable covers to finite ones, but not arbitrary ones), like $\omega_1$ in the order topology. For metric spaces all these notions (compact, countably compact, sequentially compact) all coincide, so the finer distinctions aren't always treated. The necessary motivating examples are quite technical (but interesting) and often require some set-theoretic tools.

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  • $\begingroup$ Thanks, I checked the linked answer and do I get this right? The set $[0,1]^{[0,1]}$ is compact but not sequentially compact mainly because the index set ($[0,1]$) is uncountable and there can be no convergent subsequence (via Cantor diagonalization). On the opposite side, the same base $[0,1]$ is used for the homeomorphism, but the index is $N$ and it is sequentially compact since the index of the set $[0,1]^N$ is now countable and is kinda 'similar' to the Hilbert cube (although the original set $[0,1]^I$ is uncountable). $\endgroup$
    – Link L
    Apr 30 '18 at 1:58
  • $\begingroup$ Sorry I meant ... there can be a sequence that has no convergent subsequence in the 2nd line above $\endgroup$
    – Link L
    Apr 30 '18 at 3:20
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    $\begingroup$ @LinkL the index set (for this proof) needs to be the size of the continuum. Merely uncountable isn’t enough, in general. I use the countable Cantor cube (as it’s called) as the index set (of the same size as $[0,1]$) for convenience of notation as I explain in the post. $\endgroup$ Apr 30 '18 at 5:01

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