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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.

MathIsNice1729
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1 Answers1

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

Robert Z
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  • What is the meaning of "$(f_n)_n$"? – Eric Towers Nov 08 '19 at 08:27
  • @EricTowers For each $n$, $f_n$ is an element of a specific sequence. $(f_n)_n$ is that sequence. It's somewhat common notation, although I don't use it much myself. – Arthur Nov 08 '19 at 08:29
  • @EricTowers It is the sequence of $f_n$ where $f_n$ is a tuple/function that takes the $n$th binary digit in the expansion of the index. – MathIsNice1729 Nov 08 '19 at 08:29
  • $f_n(x)$ is a bit. $f_n$ is a function. $(f_n)_n$ is a sequence of functions indexed by the naturals, so is very much not an element of ${0,1}^{[0,1]}$. Perhaps you mean $(f_n(x))_x$? – Eric Towers Nov 08 '19 at 08:31
  • @EricTowers ${0,1}^{[0,1]}$ is the set of functions $f:[0,1]\to {0,1}$. Am I wrong? – Robert Z Nov 08 '19 at 08:35
  • @RobertZ : Sure. But $(f_n)_n$ is not indexed by $[0,1]$. – Eric Towers Nov 08 '19 at 08:36
  • $(f_n)_n$ is the sequence of $f_n$ indexed by $n$. Do you prefer the notation ${f_n}_n$ ? – Robert Z Nov 08 '19 at 08:38
  • You already established that $n \in \Bbb{N}$ and have given no map $\Bbb{N} \rightarrow [0,1]$, so you have written $(f_n(?))_{n \in \Bbb{N}}$, which is not a $[0,1]$-indexed sequence. Also, relevant to the current edit: sets don't have subsequences. – Eric Towers Nov 08 '19 at 08:40
  • Why should I give a map $\Bbb{N} \rightarrow [0,1]$? A sequence in ${0,1}^{[0,1]}$ is a sequence of functions $f_n:[0,1]\to {0,1}$ with $n\in\mathbb{N}$. – Robert Z Nov 08 '19 at 08:45
  • @EricTowers Each of the $f_n$ are elements of ${0,1}^{[0,1]}$. Which is to say, each $f_n$ is a $[0,1]$-indexed sequence (thought of as a function). Then we have a sequence of those, indexed by $n\in \Bbb N$, because we want to talk about sequential compactness of ${0,1}^{[0,1]}$. – Arthur Nov 08 '19 at 08:47
  • "$(f_n)_n$" is not indexed by a variable ranging over $[0,1]$. You have established that $n$ ranges over $\Bbb{N}$. $\Bbb{N} \neq [0,1]$, but you have now repeatedly claimed that $\Bbb{N} = [0,1]$. – Eric Towers Nov 08 '19 at 08:49
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    @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. – Arthur Nov 08 '19 at 08:50
  • I have no confusion about the meaning of the second "$n$" in $(f_n)_n$ it indicates which of the two symbols in parentheses is the index. Apparently, it's the subscript on $f$ which is already established to only range over $\Bbb{N}$. – Eric Towers Nov 08 '19 at 08:52
  • @EricTowers Yes. Because each $f_n$ itself is a $[0,1]$-indexed sequence, thought of as a function. – Arthur Nov 08 '19 at 08:52
  • Where does this uncountable index appear in "$f_n$"? – Eric Towers Nov 08 '19 at 08:53
  • $\mathbb{N}\times [0,1]\ni (n,x)\to f_n(x)\in{0,1}$ – Robert Z Nov 08 '19 at 08:53
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    @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. – Arthur Nov 08 '19 at 08:53
  • So the correct notation is $(f_n(x))_x$, because $x$ actually can range over $[0,1]$. – Eric Towers Nov 08 '19 at 08:54
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    @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 ${0,1}^{[0,1]}$. – Arthur Nov 08 '19 at 08:56
  • There is nothing in "$f_n$" that can take any value in $[0,1]$. Consquently, you have not written a sequence indexed by $[0,1]$, which is what you keep saying you have written. – Eric Towers Nov 08 '19 at 08:57
  • @EricTowers It's a function. You usually don't write a variable when you write the name of a function. How many times do I have to repeat myself? – Arthur Nov 08 '19 at 08:58
  • Until you actually write an expression indexed by a variable that can range over $[0,1]$ so that can actually be an element of the set it is supposed to be. – Eric Towers Nov 08 '19 at 08:59
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    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 $(x_n(i))_i$. – Theoretical Economist Nov 08 '19 at 08:59
  • "$(f_n)n$" is literally indexed by $n$. $n$ does not range over $[0,1]$. You seem to be writing an element of $\Bbb{R}^n$ in the form $(x_i){i \in {0,1}}$. – Eric Towers Nov 08 '19 at 09:02
  • Let's come at this from the other side. What's the cardinality of what you have currently written, "${f_n}_n$"? Are there nearly enough things in that set to have an injection from $[0,1]$? – Eric Towers Nov 08 '19 at 09:06
  • A sequence in ANY topological space is indexed by natural numbers https://en.wikipedia.org/wiki/Sequence Here we have a sequence of functions ${f_n}_{n\in\mathbb{N}}$ – Robert Z Nov 08 '19 at 09:07
  • An element of ${0,1}^{[0,1]}$ is indexed by $[0,1]$. – Eric Towers Nov 08 '19 at 09:08
  • The term "sequence" is related that the fact that the space $X={0,1}^{[0,1]}$ is not "sequentially" compact https://en.wikipedia.org/wiki/Sequentially_compact_space – Robert Z Nov 08 '19 at 09:11
  • An element $x$ of $\mathbb R^n$ is a collection of members of $\mathbb R$ indexed by members of ${1,\ldots,n}$. A sequence in $\mathbb R^n$ is written as $(x_n)_n$. Similarly, an element $f$ of ${0,1}^{[0,1]}$ is a collection if members of ${0,1}$ indexed by members of $[0,1]$. A sequence in ${0,1}^{[0,1]}$ is written as $(f_n)_n$. – Theoretical Economist Nov 08 '19 at 09:11
  • @TheoreticalEconomist : "A sequence $(x_n)_n$" written after $n$ has already been bound as the dimension of the space is gibberish. – Eric Towers Nov 08 '19 at 09:15
  • I give up. Your decision to replace sequences with sets reduces readability, but that is not my main concern. Your argument is unintelligible to me after 45 minutes of trying to find a way to interpret the types to be sensible. Repeatedly claiming that a countable index somehow ranges over $[0,1]$ is not something I will ever accept. – Eric Towers Nov 08 '19 at 09:17
  • Sorry, my mistake. I should have said $\mathbb R^k$. However, I’m certain Robert Z did not at all claim that a countable index ranges over $[0,1]$. I’m sorry the explanations here have not been sufficient. However, this truly is a standard piece of notation, so I’m not sure why you think it’s mistaken. – Theoretical Economist Nov 08 '19 at 09:25
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    @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. – Arthur Nov 08 '19 at 09:56