In Rudin's principles of mathematical analysis in chapter 2, it says that the integers are countable as a set because of having the same cardinality as the positive integers. However it's just not making sense because, as the paper counts out the integers $0, 1, -1, 2, -2...$ I think of adding in $\infty$ and - $\infty$. In the positive integers to match that I add in $\infty$ plus an ordinal number higher than that one; that induces another step of adding another ordinal to the integers and the positive integers end up with 1 more cardinality. What is the mistake?
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3I'm a bit confused by the question, but $\infty$ and $-\infty$ are not integers (nor do ordinal numbers come into play, really) - hence why Rudin would not consider them. – Milo Brandt Feb 17 '20 at 05:03
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2same cardinality means they can be put in one-to-one correspondence – J. W. Tanner Feb 17 '20 at 05:04
1 Answers
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The set of integers is numerable because you can put it into a one to one correspondence. For instance you can
- $0\to 0$
- $1\to +1$
- $2 \to-1$
- $3 \to +2$
- $4 \to -2$
and so on.
Luca Goldoni Ph.D.
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I have just tried reading it again and I lost track, because there isn't infinity this time. If there are twice as many integers than positive integers, how is there a 1 : 1 correspondence when the terms seemingly run out? – Feb 17 '20 at 07:25
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@LolFlo Where do they run out? The 1-1 correspondence in this answer shows that there aren’t twice as many. – amd Feb 17 '20 at 07:33
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For example, how is counting possible after counting the positive numbers first? – Feb 17 '20 at 07:36
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I'll rephrase - picture the negative integers have cardinality equal to the positive integers. The positive integers also have that cardinality. It doesn't add up for me! – Feb 17 '20 at 09:18
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it approaches different infinities or the same infinity in either direction so im a lil confused today - _-. – Feb 17 '20 at 11:50
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Yes, the integers, the positive integers and the negative integers all have the same cardinality. This does "add up"; it simply says $\aleph_0+\aleph_0=\aleph_0$. – David C. Ullrich Feb 17 '20 at 13:06