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Let $B:=\{(b_1, b_2, b_3, \ldots) : b_i =\pm i!$ for every $i \in \mathbb{N}\}$.

I WTS that $B$ is uncountable. I know there are several ways to do this. At this point I think that constructing a surjective function from $B$ into $[0,1]$ is the easiest way. However, this should require using decimal (maybe binary) expansions of the reals and I'm unsure of how to do work with these formally.

Therefore, it would be helpful to see what such a surjection would look like and why it works.

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    Is the ! (factorial) a part of this? – 2017-02-27
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    Do you already know that the set of infinite binary sequences is uncountable? That's a much easier thing to project this onto . . . – 2017-02-27
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    Also it might be easier for you to create an injective function from [0,1] into B. – 2017-02-27
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    @NoahSchweber Yes I can use that fact. – 2017-02-27
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    @QthePlatypus I hadn't thought of that, but if that's easier I'd be interested to see that instead. Also, yes the factorial is part of it as described. – 2017-02-27
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    @CuriousKid7 OK, do you see how to assign an infinite binary sequence to each sequence of the type you're looking at? – 2017-02-27
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    @NoahSchweber Oh, I think that $f(x) =\sum_{\mathbb{N}} \frac{.5(1+\frac{b_i}{i!})}{2^i}$ works. – 2017-02-27
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    What is $i!$. if each term can only be only of two values $-i!$ or $+i!$ then this is complete equivalent to $b_i = 0|1$ which should be a standard and well known proof. – 2017-02-27
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    @fleablood What does $0\mid 1$ mean here? – 2017-02-27
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    I meant it to meant $b_i$ may be either 0 or 1. Or one of any two values. What does $i!$ mean? If there are only two values; a positive and a negative. It almost goes without saying that its the same thing as choosing 0 or 1 or any two values. The set of all sequences of 1s and 0 is called $\{0,1\}^{\infty}$ and it should be obvious that it is 1-1 with $\{$any set of two elements$\}^{\infty}$. And it is well known to be uncountable via Cantor's diagonal. – 2017-02-27
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    Oh, $i$ the natural number number index! factorial, not the the square root of negative one factorial. Okay it's misleading but it is still the same. There are two choices for each $i$. So map $\{.... b_n = \pm n! ...\} $ to $\{..... c_n...\}$ where $c_n = 0$ if $b_n < 0$ and $c_n = 1$ if $b_n > 0$. (and we can map $\{.... c_n ...\}$ to $b_n$ where $b_n = n!$ if $c_n = 1$ and $b_n = -n!$ if $c_n = 0$. That's "clearly" (I hate to use that word) 1-1 and the set of $\{c_n\}$ are known not to be countable. – 2017-02-27
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    Or you could use cantors diagonal directly. Let $\{s_n\}$ be a countable list of sequence where each sequence is $s_{n} = \{s_{n,1},s_{n,2},...\}.$. Create the sequence $t =\{-1*s_{1,1}, -1*s_{2,2}, ..... \}$. $t $ is not on the list so the list wasn't complete so a countable list is impossible so this set of sequences it uncountable. – 2017-02-27
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    "At this point I think that constructing a surjective function from B into [0,1] is the easiest way." Whoa! that is *not* easy. It's doable ... and I guess not *that* hard but way easy to map it to $\{0,1\}^{\infty}$. Way, way easier. – 2017-02-27
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    @fleablood Ok, you're right. One thing: When you say 1-1, do you mean bijective (rather than injective)? – 2017-02-27
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    I meant bijective in this case. You're right, I used it incorrectly. I always feel one to one ought mean each distinct *one* element in the mapped from set to each distinct *one* element of the mapped to set. But thats 1-1 *and* "onto". This is purely psychological but when I hear "bijection" I think too hard about the actual mechanics and rules of mapping whereas when I hear 1-1 (and onto) I of simply the elements being equivalent. I wanted to evoke the latter. Mapping {0,1} to {black, white} to {head, tail} etc. should be obvious and go without saying. – 2017-02-27
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    Of course, I completely misunderstood what $i!$ meant. That $i! \ne j!$ is a slight kink to the problems, but the only variation allowed in the $i-th$ terms is whether it is positive or negative it easy to work around. – 2017-02-27

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