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Does there exist a bijection between empty sets?

What I think is:

Since 'for every $x\in \emptyset$, there exists $y\in \emptyset$ such that $(x,y)\in f$' is false and the negation of this statement is also false because of 'existence' sentence..

But I think empty sets are equipotent intuitively.
Am I wrong? Help

Arturo Magidin
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Katlus
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    Try to use $f=\emptyset$. Is this a function from $\emptyset$ to $\emptyset$? Is it injective/surjective/bijective? (Just try to write down the definitions, it should be easy.) \ If you still have problems, maybe it would be useful if you have a look at empty function and vacuous truth at Wikipedia. – Martin Sleziak May 04 '12 at 13:04
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    "For every $x$ in $\varnothing$, ..." is always vacuously true. Kind of like being innocent until proven guilty. –  May 04 '12 at 13:13
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    There is only one empty set. So they're clearly in bijection. – lhf May 04 '12 at 13:20
  • @rahul 'for every x,y in $\emptyset$ , $\emptyset(x) is not equal to \emptyset(y)$' is true then? – Katlus May 04 '12 at 13:29
  • Consider the identity function which maps every element to itself. Since there are no elements, this is the empty function. – Mark Bennet May 04 '12 at 13:31
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    @Katlus Every claim which has the form $(\forall x\in\emptyset) \varphi(x)$ is true; where $\varphi(x)$ can be arbitrary formula. The claim that you wrote in your comment has this form. – Martin Sleziak May 04 '12 at 13:32

3 Answers3

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Yes, there is.

(As said in the comments) the function $f=\emptyset$ is the bijection you seek.

It may seem strange that $f$ is a set, but this is ok since the definition of a function is a set of pairs (s.t ...), in this case we take $f$ to be an empty set.

The definition of $f$ is fine since there are no $x, y_1,y_2 \in \emptyset$ s.t $y_1\neq y_2$ and s.t $(x,y_1),(x,y_2)\in f$.

$f$ is injective since there are no two different elements in $\emptyset$ that $f$ maps to the the same element (this is just because there are no two different elements in $\emptyset$).

$f$ is surjective since for each $y$ in $\emptyset$ there exist a source (this is since there are no elements in $\emptyset$ hence we can't say there exist an element that doesn't have a source).

Martin Sleziak
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Belgi
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Note that there are not "empty sets", because the axiom of extensionality tells us that all empty sets are equal, as they have the same elements.

Now it is obvious that there is a bijection of a set to itself. The identity, furthermore if you think of it a little bit you will see that the empty set itself is a function whose domain and range are empty, and therefore it is a bijection between that set and itself.

Further reading:

  1. Why is an empty function considered a function?
Community
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Asaf Karagila
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  • So 'if there exists a bijection between A and emptyset, A=emptyset' is false? – Katlus May 04 '12 at 13:35
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    @Katlus: What? How did you derive that from my answer? There is a bijection between the empty set and itself. Note that any bijection between two sets preserves cardinality and the only set of cardinality zero is the empty set. If there exists a bijection between $A$ and the empty set it has cardinality zero and therefore empty. – Asaf Karagila May 04 '12 at 13:37
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If a statement starts with '$\forall x \in \emptyset$,...', it is vacuously true.

Because the set contains no elements, any statement holds for 'all elements' in it.

I'm not sure if this answer is 'mathematical' enough, but I think that's the point you might have missed.

Konstantin Schubert
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    The term is "vacuously true" is more common (and has a clearer meaning, while "trivially true" implies just that the argument is "trivial"). – Asaf Karagila May 04 '12 at 13:35