Can we say the set of natural numbers is homeomorphic to the set of integers? A map $f$ from $N$ to $Z$ defined $f(n)=-n$ does not work. Could you give me any hint?
Existence of a homeomorphism between two subspaces of real line
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general-topology
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0Are you looking for a bijection or a homeomorphism? If the latter, you need to specify a topology of sorts... – 2012-10-29
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HINT: Any bijection works, since both sets have the discrete topology. Try matching up the even natural numbers with the non-negative integers and the odd natural numbers with the negative integers, for instance.
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0there exists no odd natural numbers? what do you mean? – 2012-10-29
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2Some natural numbers are very odd... – 2012-10-29
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0@ege: Of course there are odd natural numbers: I even mentioned them! You want a bijection from $\Bbb N$ to $\Bbb Z$; I’m suggesting that you try to find one that sends every even natural number to a non-negative integer and every odd natural number to a negative integer. Look at $0,1,2,3,4,5,\dots$ and $0,-1,1,-2,2,-3,\dots$ for more inspiration. – 2012-10-29
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0what about the set of rationals? Is it homeomorphic to the set of integers? – 2012-10-29
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0@ege: Definitely not. $\Bbb N$ is discrete: every singleton $\{n\}$ is an open set. Is $\{q\}$ ever an open subset of $\Bbb Q$? – 2012-10-29
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0If the topology on the rationals is standard topology,can we say is it homeomorphic to the set of integers or reals with respect to the standard topology? – 2012-10-29
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0@ege: No, definitely not. I just explained why it can’t be homeomorphic to the integers. It can’t be homeomorphic to the reals simply because $\Bbb Q$ is a countable set, and $\Bbb R$ isn’t: there isn’t even a bijection between them, let alone a homeomorphism. – 2012-10-29
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0@BrianM.Scott does this require the set N to contain {0} ? – 2017-06-10
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Hint: Natural numbers can be even or odd. Integers can be positive or negative.
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0sorry because of the misunterstanding. thanks, now its clear. – 2012-10-29
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Try $\phi(n) = (-1)^n \lfloor \frac{n}{2} \rfloor$. Then $\phi:\mathbb{N} \to \mathbb{Z}$ is a bijection.