Proof that $[0,1]^{\Bbb R}$ is not sequentially compact Can anyone please help me with this proof:

Prove that  $[0, 1]^\mathbb{R}$, where $[0, 1]$ is the unit interval in the standard topology, is NOT sequentially compact.

I need to create a sequence which has a converging subsequence, I think it is compact and therefore it is countably compact and limit point compact, correct me if I am wrong.
Thank you 
 A: An observation which makes my life easier: the infinite power like $[0,1]^{\mathbb{R}}$ (in the product topology) only "depends" on the size of the index set: if $\phi: I \rightarrow J$ is a bijection then $[0,1]^I$ is homeomorphic to $[0,1]^J$ by "shuffling coordinates" : for $f \in [0,1]^I$ define $h(f) \in [0,1]^J$ (recall that elements of the power are just functions from $I$ or $J$ into $[0,1]$) by $h(f)(j) = f(\phi^{-1}(j))$, so that $\pi_j \circ h = \pi_{\phi^{-1}(j)}$. The last identity shows that $h$ is continuous (the compositions with projections are continuous) and $h$ has an obvious inverse $\hat{h}(f)(i) = f(\phi(i))$ for all $i \in I, f \in [0,1]^J$, also continuous for the same reasons.
So, as $|\mathbb{R}| \simeq |\{0,1\}^\mathbb{N}|$, we might as well use the latter set as the index set for the power.
Define a sequence $f_n \in [0,1]^{\{0,1\}^\mathbb{N}}$ by $f_n(\omega) = \omega_n$, where $\omega \in \{0,1\}^\mathbb{N}$,( so just a sequence of $0$'s and $1$'s). This sequence has no convergent subsequence by a standard diagonal argument: 
Suppose it has a convergent subsequence, so that there are $n_1 < n_2 , <\ldots n_k < \ldots$ in $\mathbb{N}$ such that there is some $f \in [0,1]^{\{0,1\}^\mathbb{N}}$ such that $f_{n_k} \rightarrow f$ as $k \rightarrow \infty$.
Because we are working in the product topology (a.k.a. the topology of pointwise convergence) this exactly means (or at least implies by continuity of projections) that
$$\forall \omega \in \{0,1\}^\mathbb{N}: f_{n_k}(\omega) \rightarrow f(\omega), \text{ as } k \rightarrow \infty$$
Now define a special sequence $\hat{\omega} \in \{0,1\}^\mathbb{N}$ as follows: 
$\hat{\omega}_{n_{2k}} = 1$ for all $k$, and $\hat{\omega}_n = 0$ for all other $n$ not of the form $n_{2k}$. 
For $\hat{\omega}$ we have that $f(\hat{\omega})$ equals the limit of $f_{n_{2k}}(\hat{\omega}) = \omega_{n_{2k}} \equiv 1$ for the subsequence $f_{n_{2k}}, k \rightarrow \infty$ so $f(\hat{\omega}) =1$ but also $f(\hat{\omega})  =\lim_k f_{n_{2k+1}}(\hat{\omega}) = \omega_{n_{2k+1}} = 0$, as a constant sequence of $0$'s. So the pointwise convergence fails for this coordinate $\hat{\omega}$, so there can be no convergent subsequence.
As an aside: we could have played the same trick using the reals as the index set and using decimal expansions (but then you'd have to be precise about ambiguous expansions (do you use 0.99999 or 1.0000) etc.) As a diagonal argument I find using the $0-1$-sequences as index set "cleaner"; we could also have used the power set $\mathscr{P}(\mathbb{N})$ as an indexing set and the "equivalent" sequence $f_n(A) = 1$ iff $n \in A$, which makes the analogy to Cantor's argument a bit more direct...)
It is quite a lot harder to show that this size of the index set (continuum, $\mathfrak{c}$) is the smallest for which this always happens : e.g. if the continuum hypothesis fails and $\aleph_1 < \mathfrak{c}$, it is consistent that $[0,1]^{\aleph_1}$ is sequentially compact, even though the index set is uncountable while $[0,1]^{\mathfrak{c}}$ will always be compact and not sequentially compact (the same argument remains valid). 
Large products also give (IMHO) natural examples of the reverse phenomenon: sequentially compact spaces that are not compact: 
Define the following subspace of $[0,1]^I$ (where $I$ is uncountable):
$$\Sigma [0,1]^I = \{f \in [0,1]^I: \sup(f) \text{ at most countable }\}$$
were $\sup(f) = \{i \in I: f(i) \neq 0\}$, so the set of functions that are $0$ almost everywhere (except on the countable support set $\sup(f)$.
This set is easily shown to be dense in $[0,1]^I$ (even finite support would do, using basic open sets), so cannot be compact (it would be closed in [0,1]^I, not dense..) but it is  sequentially compact: suppose $(f_n)$ is a sequence in $\Sigma[0,1]^I$, define $J = \bigcup_n \sup(f_n) \subset I$, which is countable as a countable union of countable sets. Note that by definition, for all $n$ and all $i \notin J$: $f_n(i) = 0$, so outside $J$ we have a constant $0$ sequence in all coordinates. Then observe that $[0,1]^J$ is homeomorphic to $[0,1]^\mathbb{N}$, the Hilbert cube, which is metrisable in the infinite product metric (remember the first remark of this post). $[0,1]^J$ being compact metrisable, is sequentially compact so there is a convergent subsequence $f_{n_k}\rightarrow f$ in $[0,1]^J$. So we have pointwise convergence to $f$ on $J$ and setting $f(i) = 0$ for all $i \notin J$, we have it for all coordinates.
So $\Sigma[0,1]^I$ is sequentially compact and not compact. So neither property implies the other in general. (they do for metric spaces as is well known).  
