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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.

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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 $\{0,1\}^{[0,1]}$.

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    $\begingroup$ @EricTowers $\{0,1\}^{[0,1]}$ is a space of $[0,1]$-indexed sequences. We are looking at $\Bbb N$-indexed sequences in that space, thus a sequence of sequences. Robert's presentation of his such sequence makes perfect sense. $\endgroup$
    – Arthur
    Nov 8 '19 at 8:50
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    $\begingroup$ @EricTowers It doesn't. Because when you write the name of a function (in this case $f_n$ for all the various $n$), you usually don't write the variable name. $\endgroup$
    – Arthur
    Nov 8 '19 at 8:53
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    $\begingroup$ @EricTowers No, the correct notation is $(f_n)_n$, because you usually don't write the variable when you write the name of a function. Writing $f_n(x)$ implies that you're talking about the function value for a specific $x$, which would make $(f_n(x))_n$ a $\Bbb N$-indexed sequence in $\{0,1\}$, not $\{0,1\}^{[0,1]}$. $\endgroup$
    – Arthur
    Nov 8 '19 at 8:56
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    $\begingroup$ Insisting that the correct notation is $(f_n(x))_x$ is like insisting that the correct way to write a sequence in $\mathbb R^n$ is $(x_n(i))_i$. $\endgroup$ Nov 8 '19 at 8:59
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    $\begingroup$ @EricTowers We're not replacing sequences with sets. We're replacing (the $[0,1]$-indexed) sequences with functions. In fact, I personally read $X^Y$ for sets $X, Y$ first and foremost as a set of functions, not of sequences. So a sequence in $X^Y$ is a sequence of functions, and writing that sequence as $(f_n)_n$ is completely natural. $\endgroup$
    – Arthur
    Nov 8 '19 at 9:56

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