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Assume we have a set $X$ which is a (closed) line segment. Prove that if we split $X$ into 2 parts $X_1$ and $X_2$ then at least one of those sets would have the same cardinality as $X$.

My attempt: 1) Let's assume that $X_1$ contains some line segment itself, let's call it $Y$.

2) We know that there is a bijection between $X$ and $Y$ and therefore $|X| = |Y|$.

3) And since $X_1$ is embedded in $X$ we can use Cantor-Bernstein theorem to conclude that $|X_1| = |X|$.

Now, I am not sure how to prove the second case: when $X_1$ (or $X_2$) does not contain a line segment. Is there a way to use Cantor-Bernstein there?

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Hint: assume both parts have cardinality less than R. We know that their sum is R. Can you derive a contradiction from A+B=R, A,B<R? –  Lopsy Jan 1 '12 at 21:23
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2 Answers 2

up vote 4 down vote accepted

Indeed some axiom of choice is needed. Monro shows in [1] that it is possible to have the continuum as the sum of two strictly smaller cardinals (He in fact shows a bit more, that it is consistent to have all the non-well orderable cardinals with this property).

Quite the bizarre behavior, but that should not surprise people familiar with the behavior of cardinals without the axiom of choice.

Assuming sufficient amount of choice, the question becomes quite simple - as Henning explains.

Since $2^{\aleph_0}=|X_1|+|X_2| = \max\{|X_1|,|X_2|\}$ it cannot be that both $X_1$ and $X_2$ have a strictly smaller cardinality.

A side remark to your approach, while it is a valid argument to divide to cases it should not be a concern here because there are a lot of ways to break the line into non-line segments. Choose a dense co-dense subset, and there are plenty of those. When working with cardinal arithmetics it is a good advice to know exactly how far your assumptions allow you to stretch "the cardinality games" before reducing into cases and construction bijections by hand. In this case, some choice is needed, and in turn we have the maximality argument of the sum.


Bibliography:

  1. Monro, G.P. Decomposable Cardinals. Fund. Math. vol. 80 (1973), no. 2, 101–104.
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Surprising. Thanks for the reference! –  Andres Caicedo Jan 1 '12 at 22:36
    
@Andres: Quite surprising! Morno got a lot of such results, and I am knee deep in his papers in the past couple of months. –  Asaf Karagila Jan 1 '12 at 22:38
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If you don't know that either of the parts contain an interval, then I think you need the axiom of choice.

In that case, $X$ is the disjoint sum of $X_1$ and $X_2$, so $|X|=|X_1|+|X_2|$. And it's a set-theoretic fact that $AC\Rightarrow \kappa+\lambda=\max(\kappa,\lambda)$ if $\kappa$ and $\lambda$ are cardinals and at least one of them is infinite.

Instead of Axiom of Choice, you could also assume the Continuum Hypothesis, which says that any subset of $\mathbb R$ that is not equinumerous with $\mathbb R$ is at most countable. Then if neither $X_1$ nor $X_2$ are equinumerous with $\mathbb R$, their union would also be at most countable, a contradiction. (But CH is generally considered a less orthodox assumption than AC, so this variant is not recommended for homework).

It is a bit strange, though, that these questions are framed in terms of "line segments" and such, rather than abstract unstructured sets. This limits them to being about the continuum rather than about a generic cardinality. Are you sure the partition into $X_1$ and $X_2$ doesn't have to respect some geometric (or topological) structure?

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