- #1
kidsmoker
- 88
- 0
Hi,
there's a proposition in my lecture notes which states "If X is a set, there can be no surjection [tex]X \rightarrow P(X)[/tex]", where P(x) is the power set of X (does anyone know how I get the squiggly P?).
The proof given there seems unnecessarily complicated to me. Would the following be okay?
We need to show that if X is a set, there can be no surjection [tex]X \rightarrow P(X)[/tex].
Suppose that [tex]p:X \rightarrow P(X)[/tex] is a map. Now P(X) is the set of all the subsets of X. Define p(x) = {x}.
This maps each [tex]x \in X[/tex] to the most obvious subset {x}. Then clearly there is the subset [tex]\emptyset \in P(X)[/tex] such that [tex]p(x) \neq \emptyset[/tex] for any [tex]x \in X[/tex]. Hence no surjection can exist.
This seems to break down when X is the empty set (in which case there is a surjection surely?) but other than that it seems okay to me.
Any help appreciated!
there's a proposition in my lecture notes which states "If X is a set, there can be no surjection [tex]X \rightarrow P(X)[/tex]", where P(x) is the power set of X (does anyone know how I get the squiggly P?).
The proof given there seems unnecessarily complicated to me. Would the following be okay?
We need to show that if X is a set, there can be no surjection [tex]X \rightarrow P(X)[/tex].
Suppose that [tex]p:X \rightarrow P(X)[/tex] is a map. Now P(X) is the set of all the subsets of X. Define p(x) = {x}.
This maps each [tex]x \in X[/tex] to the most obvious subset {x}. Then clearly there is the subset [tex]\emptyset \in P(X)[/tex] such that [tex]p(x) \neq \emptyset[/tex] for any [tex]x \in X[/tex]. Hence no surjection can exist.
This seems to break down when X is the empty set (in which case there is a surjection surely?) but other than that it seems okay to me.
Any help appreciated!