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December 25th, 2016, 06:18 PM   #1
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How are integers a subset of R^2?

I am reading an Analysis book and it says, "Let us consider the following subsets of $\displaystyle R^2$"

It goes on to mention these sets:

The set of all integers

The set consisting of the numbers $\displaystyle \frac 1n$ for $\displaystyle (n=1,2, 3,...)$

The segment $\displaystyle (a,b)$

My question is, isn't this $\displaystyle R^1$? How is this $\displaystyle R^2$?

It may seem obvious, but I really don't know.. Thanks for your help!

Last edited by ProofOfALifetime; December 25th, 2016 at 06:20 PM.
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December 25th, 2016, 06:33 PM   #2
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Surely you agree that the integers are a subset of the reals? An element of $\mathbb{R^1}$ could be (a), and an element of $\mathbb{R^2}$, (b, c). Does it not follow that the integers are also apart of the $\mathbb{R^2}$ space?
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December 25th, 2016, 09:32 PM   #3
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I must say that's a strange way to put things, and I wonder if you've read the book correctly

It's true that the set $\{(n, 0)\} \in \mathbb R^2 : n \in \mathbb Z\}$ is commonly identified with the integers, so that we would typically say that the integers are a subset of $\mathbb R^2$ without worrying that this is not strictly true in terms of set theory. In this case we implicitly invoke a structuralist position that anything isomorphic to the integers may be called the integers. There is actually some math philosophy in there; because this example shows that there is something not quite right about using set theory to model mathematical objects. https://en.wikipedia.org/wiki/Struct...of_mathematics)

In any event, it would be very unusual for an analysis text to refer to the integers as a subset of the plane in the manner you indicate; and likewise your other examples, especially $\{\frac{1}{n}\}$.

I wonder if you can double check your book and if possible, post a scan or photo of the page in question.
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December 26th, 2016, 06:19 AM   #4
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It's Rudin's book Principles of Mathematical Analysis. Anyways, here it is.
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December 26th, 2016, 06:23 AM   #5
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I have always learned that the integers are a subset of the $\displaystyle R^1$. So are we thinking of them as $\displaystyle (n,0)$? Or as $\displaystyle (a,b)$? Like, I said, it just didn't seem clear from the text.

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December 26th, 2016, 06:26 AM   #6
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Thank you both for your answers, I'm sorry if the question isn't phrased well or if it's too obvious.
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December 26th, 2016, 08:35 AM   #7
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He explicitly says, "Note that (d), (e), and (g) are also subsets of $\mathbb R^1$". And later, he says he's leaving an entry blank because the answer changes depending on whether you consider an open interval to be in $\mathbb R^1$ or $\mathbb R^2$. Those phrases remove the ambiguity. In the other cases the answers don't depend on which space you're considering.
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December 26th, 2016, 08:45 AM   #8
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That doesn't answer my question, the question was, how are those sets viewed as subsets of $\displaystyle R^2$?

I guess it just means elements like $\displaystyle (a,b)$where $\displaystyle a$ and $\displaystyle b$are integers.

And $\displaystyle (\frac 1n, \frac 1n)$ where $\displaystyle n=1,2,3...$

I just was hoping for some clarification, not just someone to repeat what is already on the page..
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December 26th, 2016, 10:57 AM   #9
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Quote:
Originally Posted by ProofOfALifetime View Post
That doesn't answer my question, the question was, how are those sets viewed as subsets of $\displaystyle R^2$?

I guess it just means elements like $\displaystyle (a,b)$where $\displaystyle a$ and $\displaystyle b$are integers.

And $\displaystyle (\frac 1n, \frac 1n)$ where $\displaystyle n=1,2,3...$

I just was hoping for some clarification, not just someone to repeat what is already on the page..
I already clarified it in my first response. The set $\{(n,0) \in \mathbb R^2 : n \in \mathbb Z\}$ is identified with the integers. Likewise we identify $\frac{1}{n}$ with $(\frac{1}{n}, 0)$.

@Joppy made the same point before I did.

Last edited by Maschke; December 26th, 2016 at 11:02 AM.
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December 26th, 2016, 01:22 PM   #10
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Actually your first response did not sound like clarification, if anything it sounded unsure. It would have helped if you mentioned in your second response that YES what you said earlier is the answer to what I was asking.

Oh well!

Last edited by ProofOfALifetime; December 26th, 2016 at 01:25 PM.
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