Show that $\{0,1\}^{[0,1]}$ is not sequentially compact
Obviously, it is taken with the product topology of the subspace topologies (which are, in fact, discrete).
Now, the elements are tuples of $0$'s and $1$'s with $[0,1]$ as the indexing set. For a sequence to converge in this topology, for any collection of finite indices, $\exists m \in \mathbb{N}$ such that the terms of the sequence match the limit in those indices after that $m$. I cannot go any further.
Hints are welcome rather than complete answers.
Best Answer
For $x\in [0,1]$ consider its unique binary expansion that has not an infinite tail of ones and let $f_n(x)$ be the $n$-th binary digit of $x$. Therefore $$x=\sum_{n=1}^{\infty} \frac{f_n(x)}{2^n}.$$ Does the sequence $(f_n)_{n\in\mathbb{N}}$ have a convergent subsequence in $\{0,1\}^{[0,1]}$?
The answer is no. Take any subsequence $(n_k)_{k\in\mathbb{N}}$ and consider the point $x=\sum_{j=1}^{\infty} \frac{1}{2^{n_{2j}}}\in [0,1].$ Then $f_{n_k}(x)=1$ when $k$ is even and $f_{n_k}(x)=0$ otherwise, which implies that the sequence $(f_{n_k}(x))_{k\in\mathbb{N}}$ in $\{0,1\}$ is not convergent and therefore $(f_{n_k})_{k\in\mathbb{N}}$ is not convergent in $\{0,1\}^{[0,1]}$.