for any set X, construct and injection from X to Power set of X this is what i think, if i assume X to be {1,2,3}
then P(x) will have {{1},{1,2},{1,2,3},{1,3},{3,2},{3,2,1},etc}}
so will not, to say, 1 from X map to more than one element of P(x) ? so how can i prove for injection ?
 A: For every $x \in X$, map $x$ to $\{x\} \in 2^X$. This is an injection.
A: Yes, the map provided by Adhvaitha is the simplest injection possible. But you should inderstand this. The map you have to construct associate to each element x of X an element (just one) of P(X). The elements of P(X) are ALL subsets of X, so you should associate to each x a subset of X (which is considered as one element of P(X)). For example you can associate to each x of X the empty set which is an element of P(X). But of course this is not injective; it is a consatante map. Adhvaitha suggested to each x to associate le subset of X which contains only the element x, that is {x}. Why this is an injection? Answer: Assume that there is x and y in X such that f(x)=f(y) then {x}={y} (because by definition f(x)={x} and f(y)={y}, remember we have said that for EACH x we associate {x} so to y we associate {y}) Now because the set {x}={y} then each element of {x} is also an element of {y} so, the unique element x of {x} is an element of {y} which contains only y, so there is no choice x has to be y, so, x=y. The injection is proved.
