Closed but not sequentially compact $I^I$ 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 $I^I$?
 A: $I^I$ (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.
