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A classic example of homeomorphism is between a sphere missing one point and a plane

To see this, place a sphere on the plane so that the sphere is tangent to the plane. Given any point in the plane, construct a line through that point from the "north" pole (assuming the point of tangency is the south pole); this necessarily intersects the sphere in a unique point other than the north pole. Similarly, given any point on the sphere other than the north pole, the line through the north pole and that point will intersect the plane at a unique point. Thus, the homeomorphism is established.

Since there is a homeomorphism between a sphere missing one point and a plane, does it make sense to say that a sphere has one more point than a plane, even though both sets are infinite and uncountable?

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A simpler example: take the function $f:\mathbf N_{\geq 0}\to \mathbf N_{>0}$, $f(n)=n+1$. $f$ is a bijection. There are as many positive natural numbers as there are natural numbers. A sphere without one point has as many points as sphere. There were a lot of questions like yours here, for example this one just yesterday. – tomasz Aug 14 '12 at 14:17
In fact, a set is infinite if and only if it can be placed in bijective correspondence with a proper subset of itself. – Charles Staats Aug 14 '12 at 14:23
@CharlesStaats: Assuming axiom of choice, that is. ;) – tomasz Aug 14 '12 at 14:40
up vote 5 down vote accepted

Since there is a homeomorphism between a sphere missing one point and a plane, does it make sense to say that a sphere has one more point than a plane, even though both sets are infinite and uncountable?

It can certainly be given meaning. For example, given any inclusion of subsets $A \subseteq B$, we can define the phrase "$B$ has $\alpha$ more elements than $A$" to mean that $|B \setminus A| = \alpha$.

More generally, given any injection $i: A \to B$, we can define "$B$ has $\alpha$ more elements than the subset defined by $i$" or "$B$ has $\alpha$ more elements than $A$ relative to the inclusion $i$" to mean $|B \setminus i(A)| = \alpha$.

These notions can even be useful. For example, the notion of a cofinite subset often comes up: a subset $S$ of a set $X$ such that $|X \setminus S| < \aleph_0$.

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It should be stressed that this really measures the size with regards to the inclusion map. Different inclusions give rise to different sizes – Thomas Rot Aug 14 '12 at 14:56

Yes, and no, it really just depends on how you're defining the "size" of sets. In each case you're applying a metric, or something close to it (since not every measure of size inherently fits the definition of a metric in a more formal sense).

I can easily say that, in the real numbers, the size of a set is it's lebsgue measure, or even the maximal distance of it's supremum and infimum; in the same way you can define one set to be larger than another if the removal of an element allows a homeomorphism to exist between them. This can then be used to create an equivalence relation of the sets in question, and an ordering between those equivalence classes.

Ordinarily cardinality is used in this sense, but it's not the only way to define a set's size.

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Potential problems: 1) There exist spaces that are homeomorphic to the same space with a point removed. 2) The order relation you have defined gives no relation at all between, say, $\mathbb N$ and $\mathbb R$. – Charles Staats Aug 14 '12 at 14:55
  • Is it true that there is a one-to-one correspondence between $B$ and $A\setminus\{a\}$, where $a\in A$, and that there is no one-to-one correspondence between $B$ and $A$?
  • Is it true that there is a well-behaved one-to-one correspondence between $B$ and $A\setminus\{a\}$, where $a\in A$, and that there is no well-behaved one-to-one correspondence between $B$ and $A$?

The first can certainly be true if $A$ and $B$ are finite. In conventional set theory, it cannot be true if they're infinite.

The second depends on which notion of "well-behaved" is invoked, and there you have a vast menu to choose from. Sometimes "well-behaved" means "linear", and then a three-dimensional space is bigger than a two-dimensional space. Sometimes well-behaved means a homeomorphism, and then you've got your plane-and-sphere example. Sometimes well-behaved means measure-preserving. Sometimes it means an isometry.

Here's an example that I like. In non-standard analysis one has "infinite integers". Let $n$ be one such. Consider the sets $\{1,2,3,\ldots,n\}$ and $\{1,2,3,\ldots,n,n+1,n+2\}$. If "well-behaved" has no content, so that all bijections are well-behaved, then these two sets have the same cardinality and that's that. But in non-standard analysis one speaks of internal and external objects. Every internal (i.e. well-behaved) one-to-one function from $\{1,2,3,\ldots,n\}$ to $\{1,2,3,\ldots,n,n+1,n+2\}$ omits exactly two members of the latter set from its image. Take your pick as to which two they will be.

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Size of sets is measured by cardinality. A set $A$ is said to have a smaller or equal cardinality than $B$ if there exists an injective function $f:A\rightarrow B$. If there exists a bijective function the sets have the same cardinality.

A nonempty set is finite if it has the same cardinality as the set $\{1,\ldots,n\}$ for some $n\in\mathbb{N}$. Otherwise it is infinity. It is countable if it is in bijective correspondence with $\mathbb{N}$

The following sets have the same cardinality, even though "intuitively" one would expect this not to be the case

Natural numbers $\mathbb{N}$, The even natural numbers $2\mathbb{N}$, the set of prime numbers $P$, The set of all integers $\mathbb{Z}$. The rationals $\mathbb{Q}$.

The set of reals $\mathbb{R}$ has a (strictly) larger cardinality than the set of natural numbers though. This is Cantors famous diagonal argument.

In the example you describe, you give an injection (but not bijection) of the plane into the sphere. But there also exists an injection (not a bijection) of the sphere into the plane.

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