Can a one-to-one mapping of a set into itself prove that the set is infinite?

[cragar]

Homework Statement

Let X be a set and let f be a one-to-one mapping of X into itself such that
$f[X] \subset X$ Then X is infinite.

The Attempt at a Solution

Let's assume for the sake of contradiction that X is finite and there is an f such that it maps all of the elements of X to a proper subset of X called Y. And this function also preserves the 1-to-1 mapping. If Y is a proper subset then |Y|<|X| and if X is finite then
Y is also finite, but there is a problem with the one to one mapping, if Y has less elements then X then there is no 1-to-1 mapping. So this is a contradiction of our original assumption therefore X is infinite.

 

[voko]

The reasoning is correct in general, but perhaps somewhat informal. Formally, you could recall that a one-to-one mapping of two sets implies that the two sets have equal cardinality. And for finite sets, cardinality is simply the number of their elements.

 

[clamtrox]

The reasoning is correct in general, but perhaps somewhat informal. Formally, you could recall that a one-to-one mapping of two sets implies that the two sets have equal cardinality. And for finite sets, cardinality is simply the number of their elements.

I think you are thinking of mappings that are both one-to-one and onto. If the mapping is only one to one, then the cardinalities don't need to be equal -- there for example certainly is a one to one mapping from N to R, f(x) = x.

 

[voko]

I think you are thinking of mappings that are both one-to-one and onto. If the mapping is only one to one, then the cardinalities don't need to be equal -- there for example certainly is a one to one mapping from N to R, f(x) = x.

No, what I mean is that the cardinality of the image of X (where f is onto) must be equal to that of X. Invoking the finiteness of X completes the proof.

 

[HallsofIvy]

No, what I mean is that the cardinality of the image of X (where f is onto) must be equal to that of X. Invoking the finiteness of X completes the proof.

"
But nothing is said about f being "onto".

 

[voko]

"
But nothing is said about f being "onto".

Any map is "onto" its image.