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Given two sets that have the same cardinal number

Example: \begin{align*} A & = \{1, 4\}\\ B & = \{1, 2\} \end{align*} How would you prove that the function from $A$ to $B$ is always injective and surjective AND not.... injective but not surjective or surjective but not injective.

My proof: Since the cardinal number of $A$ and $B$ is $n(A) = n(B)$. Then thus, inj$(\beta) \wedge$ surj$(\beta)$. Therefore by definition, bij $(\beta)$. So the function from $A$ to $B$ is always bijective.

Is this correct? Other ways to prove this?

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closed as unclear what you're asking by Aloizio Macedo, N. F. Taussig, Empty, Claude Leibovici, Alex M. Oct 6 '15 at 7:25

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  • $\begingroup$ Sets are not injective, functions are injective. Do you mean that any function that is injective or surjective function $A\to B$ is necessarily both injective and surjective? $\endgroup$ – Thomas Andrews Oct 5 '15 at 20:10
  • $\begingroup$ This sounds confused. Two sets have the same cardinality given that one can show that there exists a bijection between them. $\endgroup$ – S Valera Oct 5 '15 at 20:12
  • $\begingroup$ @ThomasAndrews Sorry. yes the function from A to B. I want to prove that if a fuction has the same cardinal number then it has to be bijective $\endgroup$ – Mathy Oct 5 '15 at 20:17
  • $\begingroup$ That it is impossible for it to be: injective and not surjective or surjective and not injective $\endgroup$ – Mathy Oct 5 '15 at 20:18
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What you're trying to prove is simply not true. First of all a function from $A$ to $B$ does not have to be injective and surjective. Take for example the set's in your example and have $f:A\to B$ such that $f(1)=f(4)=2$, clearly not injective or surjective.

Also it's possible for an injection to not be surjective and a surjection to be not injective. For the first case just take the mapping from $\mathbb Z$ to $\mathbb R$ such that $f(x) = x$, clearly an injection that is not a surjection. The other way around from $\mathbb Z$ to $\mathbb N$ such that $f(n) = |n|$, a surjection that's not injective.

What you could try to prove is that for finite sets with the same cardinality that a map $f:A\to B$ is injective is equvialent to it being surjective.

But if a mapping $f:A\to B$ is injective we have that $|A| \le |f(A)|$, but $f(A)\subseteq B$ so $|f(A)|\le|B|$ so we have:

$$|A| \le |f(A)| \le |B| = |A|$$

And since $|f(A)| = |B|$, $f(A)\subseteq B$ and $B$ is finite we must have that $f(A) = B$, that is $f$ is surjective.

The opposite is proved in similar way.

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  • $\begingroup$ My question was terribly reworded but you were able to answer it. Thanks. $\endgroup$ – Mathy Oct 5 '15 at 21:27

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