# Does there exist a bijection between empty sets?

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

• 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. May 4, 2012 at 13:04
• "For every $x$ in $\varnothing$, ..." is always vacuously true. Kind of like being innocent until proven guilty.
– user856
May 4, 2012 at 13:13
• There is only one empty set. So they're clearly in bijection.
– lhf
May 4, 2012 at 13:20
• @rahul 'for every x,y in $\emptyset$ , $\emptyset(x) is not equal to \emptyset(y)$' is true then? May 4, 2012 at 13:29
• @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. May 4, 2012 at 13:32

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).

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.

• @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. May 4, 2012 at 13:37
If a statement starts with '$\forall x \in \emptyset$,...', it is vacuously true.