Set and cardinality injection and surjection proof

Question

Let X be a set. Prove there is an injection from $X \rightarrow 2^X$. Prove that there is not a surjection from $X \rightarrow 2^X$.

My try- Assume to the contrary that $f: X \rightarrow 2^X$. is a surjection and consider the set $M=\{x\in X | x \not\in f(x)\} $. Then show to show that M doesn't have a pre-image. So, there is no surjection.

You wanted to show it is an surjection. Why do you want to show "there is an infinite number of sizes of infinity cardinals". The last part of your question is not clear. — Babai, Jul 21 '14 at 19:29

score 0 · Answer 1 · answered Jul 21 '14 at 19:21

0

hint: For injection consider the function $f:X \longrightarrow 2^X$ given by $f(a)=\{a\}$.

answered Jul 21 '14 at 19:21

Anurag A

41,067

score 0 · Answer 2 · answered Jul 21 '14 at 19:45

Let us M has a pre-image. Let us say $f(y)=M$.

Either $y\in M$ or $y\not \in M$

Now, if $y\in M $ then $y\not \in f(y)$ (by construction of M). But $f(y)=M$. So, $y\not \in M$. So a contradiction.

If $y\not \in M $ then $y \in f(y)$(by construction of M) . But $f(y)=M$. So, $y \in M$. So a contradiction.

Therefore M doesn't have a pre-image. That implies there is no surjection from $X \rightarrow 2^{X}$

Set and cardinality injection and surjection proof

2 Answers2