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I've been having a crack at my university's Christmas challenge (link at the bottom) and my solution to Q2 relies on me creating a bijective function and 2 sets $A$ and $B$ such that:

$A \cup B = \mathbb{R}$

$A \cap B = \emptyset$

$|A| = |B|$ (implied by the bijective function, I think)

As much as I'd like someone to simply provide an answer, I'm more interested in how to actually create one myself. Thanks in advance!

http://www.lboro.ac.uk/media/wwwlboroacuk/external/content/schoolsanddepartments/mathematicalsciences/documents/ChristmasChallenge2016.pdf

Ben Crossley
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1 Answers1

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Note first that any nonempty interval has the cardinality of $\mathbb{R}$, so any subset of $\mathbb{R}$ which contains a nonempty interval must have cardinality $|\mathbb{R}|$.

So you just need $A$ and $B$ to contain an interval each, and to be disjoint, and to cover $\mathbb{R}$.

The "obvious" answer is to take the positive numbers and the nonpositive numbers, but I'm afraid I don't know how I came up with that; I just saw it.

That yields a slightly messy bijection. If you want an extremely easy bijection, try $$A = \bigcup_{n \in \mathbb{Z}} [2n, 2n+1), B = \bigcup_{n \in \mathbb{Z}} [2n-1, 2n)$$ and $f(x) = x+1$.

Patrick Stevens
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  • But is there a bijection between those two sets? – jnyan Dec 11 '16 at 08:19
  • @jnyan Of course. The first two sentences explain why they have the same cardinality. (There's several easy ways to see that there's an explicit bijection anyway. Cantor-Schroeder-Bernstein is one way, or there's the standard trick for finding a bijection between infinite $A$ and $A \cup {x}$: select a countable subset $x_1, x_2, \dots$ of $A$, and make the function $f: x_i \mapsto x_{i+1}$, $x \mapsto x_1$, and $a \mapsto a$ for any $a$ which is not $x$ or an $x_i$.) – Patrick Stevens Dec 11 '16 at 08:21
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    I think you mean the positive and the nonpositive numbers. @jnyan: Yes, basically you just have to modify a bit the function $f(x)=-x$ to handle the problem with $0$. – Redundant Aunt Dec 11 '16 at 08:23
  • I don't see how. Please help me out. Continous function maps inverse of open sets to open sets. – jnyan Dec 11 '16 at 08:25
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    @jnyan "Continuous" was not specified in the question. – Patrick Stevens Dec 11 '16 at 08:26
  • But x+1 is a continuous function. – jnyan Dec 11 '16 at 08:28
  • (For context, I added $x \mapsto x+1$ some time after jnyan's first and second comments, but before the third.) It is true that the OP's question can be done even with the added constraint of continuity, but that's not what they asked for. – Patrick Stevens Dec 11 '16 at 08:29
  • @jnyan Also your definition of "continuous" is a little inaccurate. A continuous function is one with the property that the preimage of each open set is open. No need for invertibility (though of course in this case we're dealing with bijections which are by necessity invertible, so it can be stated as "the image of each open set under the inverse of the function is open"). – Patrick Stevens Dec 11 '16 at 08:32