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Here, Terence Tao writes:

… by using the decimal representation to embed ${2^{\bf N}}$ into ${{\bf R}}$.

What does he mean by "embed"? Is he speaking about constructing an injection ${2^{\bf N}}$ into ${{\bf R}}$? Is says using the "decimal representation"? Shouldn't it be binary representation?

How to "embed" ${2^{\bf N}}$ into ${{\bf R}}$?

Please Help
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  • Why should it be the binary representation? – Improve Feb 21 '17 at 17:39
  • I suppose he means $2^{\mathbb{N}}$ as the powerset of $\mathbb{N}$. – Fimpellizzeri Feb 21 '17 at 17:40
  • Yes, he is talking about an injection for $2^N$ inte $R$. Can you do that? What is an element of $2^N$? – fleablood Feb 21 '17 at 17:45
  • He uses decimal notation, because in binary notation the map would no longer be injective. – Mark Feb 21 '17 at 17:49
  • I've added a way to make things injective in binary. – Noah Schweber Feb 21 '17 at 17:49
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    @Fimpellizieri He might be. But he could also be talking about $2^N$ as the set of all sequences (finite and infinite) of ${0,1}$. There's a very natural injection from that to R if you consider a binary decimal expansion. If the poster remembers the injection from P(N) to {sequences of {0,,1}} that can work too but it has a "translation" extra step. – fleablood Feb 21 '17 at 17:50
  • @fleablood: (interpreting as a binary numeral gives a very natural map, but it's not an injection) –  Feb 21 '17 at 18:08
  • In analysis, the image of the embedding is often called a Cantor set. https://en.wikipedia.org/wiki/Cantor_set – Carl Mummert Feb 21 '17 at 18:14
  • Not a binary numeral. A binary decimal. Which you are right is not injective as .01111111111111111 = .1. Which I didn't consider because it is fixable. But simply into decimals with only one and zeros is definitely injective (but not surjective. – fleablood Feb 22 '17 at 00:16

2 Answers2

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He is indeed talking about constructing an injection. However, it's even better than that: $2^\mathbb{N}$ carries a natural topology - even a metric! - and according to that, the injection Tao mentions is continuous.

You ask about using binary vs. decimal - there is a slight issue in using binary: reals with multiple binary expansions. E.g. in binary $0.0111111...=0.100000...$, so the "obvious" map is not injective. Things are easiest in base $>2$ - decimal, or ternary, or whatever you want.

One way to do things via decimal notation would be as follows:

  • Take your binary sequence,

  • change each $1$ to a $5$,

  • and put a decimal point at the end.

E.g. "$0, 1, 1, 0, 1, 0, 0, . . .$" becomes "$0.0550500...$". This is indeed injective, as is easily checked. And there are lots of other ways to do this, too. Tao says to use decimal notation just for simplicity - if you prefer to work in a different base you can, but if you work in binary you need to be careful to ensure injectivity.

One way to do this in binary is to use a different replacement scheme - e.g. replace each "$0$" with "$01$", and each "$1$" with "$10$", so you're using more than one digit. Then e.g. the sequence "$01001101...$" turns into the number $0.0110010110100110...$. This is less elegant, though, than just using a different base.

Noah Schweber
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  • Moreover, because $2^\mathbb{N}$ is compact, a continuous injection from $2^\mathbb{N}$ is also an open map onto its range, so it gives a homeomorphism between $2^{\mathbb{N}}$ and a subset of $\mathbb{R}$. – Carl Mummert Feb 21 '17 at 17:52
  • Noah, can you answer this question: http://math.stackexchange.com/questions/2155108/is-most-of-mathematics-independent-of-set-theory It is about something you once said. – Please Help Feb 21 '17 at 19:24
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An element of $\{0,1\}^{\mathbb N}$ is a sequence $(a_0, a_1, a_2, \ldots)$ where each $a_i$ is either $0$ or $1$. Just build a real number out of them.

For instance, the sequence $(1, 0, 0, 1, 1, 1,\ldots)$ is mapped to the real number $0.100111\ldots$. This mapping is injective, i.e., an embedding.

Mark
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