MathematicalPhysicist

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## Main Question or Discussion Point

i have this question:

if: X=U X_i for every i in I, where X_i's are non empty and are disjoint, then |X|>=|I|.

obvously there's the one to one function from I to X, which is g(i)=X_i, but my question is according to my text i need to use here the axiom of choice, i dont think i used here the axiom of choice, is there another way to prove this statement with the axiom of choice?

and also im not sure my approach is correct cause X_i is a subset of X and not an element in X.

perhaps g is a one to one from I to P(X)\{empty set}, and from the axiom of choice we are assurd that there exists a function f:P(X)\{empty set}->X so if we take fog we have a function from I to X which is one to one cause g is one to one:

fog(i)=fog(j)=> f(X_i)=f(X_j) and f(X_i) is in X_i and f(X_j) is in X_j for every X_i in P(X)\{empty set}, so X_i and X_j arent disjoint which means that i=j.

am i correct here?

if: X=U X_i for every i in I, where X_i's are non empty and are disjoint, then |X|>=|I|.

obvously there's the one to one function from I to X, which is g(i)=X_i, but my question is according to my text i need to use here the axiom of choice, i dont think i used here the axiom of choice, is there another way to prove this statement with the axiom of choice?

and also im not sure my approach is correct cause X_i is a subset of X and not an element in X.

perhaps g is a one to one from I to P(X)\{empty set}, and from the axiom of choice we are assurd that there exists a function f:P(X)\{empty set}->X so if we take fog we have a function from I to X which is one to one cause g is one to one:

fog(i)=fog(j)=> f(X_i)=f(X_j) and f(X_i) is in X_i and f(X_j) is in X_j for every X_i in P(X)\{empty set}, so X_i and X_j arent disjoint which means that i=j.

am i correct here?