# Show that $\{0,1\}^{[0,1]}$ is not sequentially compact

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.

• A slight variation on the argument by @RobertZ below is given in my answer here . The size of the domain $I$ or $[0,1]$ is the same so the spaces are homeomorphic. Nov 8 '19 at 20:31
• Does this answer your question? Product space that is compact, but isn't sequentially compact May 2 '20 at 22:55

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 $$\{0,1\}^{[0,1]}$$.
• @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. Nov 8 '19 at 8:50
• @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. Nov 8 '19 at 8:53
• @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 $\{0,1\}^{[0,1]}$. Nov 8 '19 at 8:56
• 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 $(x_n(i))_i$. Nov 8 '19 at 8:59
• @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. Nov 8 '19 at 9:56