Let A be a subset of X. Suppose there is an injective function from X to A...

Let X be a set and let A be a subset of X. Suppose there is an injection $f: X \rightarrow A$. Show that the cardinalities of A and X are equal.

I tried proving this for hours but couldn't really get anywhere. So I was wondering if anybody could give me a hint so that I could start from there. By the way, I'm not supposed to use the Cantor-Bernstein Theorem.

Thanks in advance

Let X be a set and let A be a subset of X. Suppose there is an injection $f: X \rightarrow A$. Show that the cardinalities of A and X are equal.

I tried proving this for hours but couldn't really get anywhere. So I was wondering if anybody could give me a hint so that I could start from there. By the way, I'm not supposed to use the Cantor-Bernstein Theorem.

Thanks in advance

I am about to either over simplify it and lose all rigour or about to make my first contribution to TSR. But here goes;

From hypothesis that A is a subset of X, don't we get |A| =< |X| ?

Next also from hypothesis that there exists f: X-->A s.t f is an injection, that |A| >= |X| ??

But both are only true when they are equal.

It's not an unreasonable thing to think that is should be true, but this is a really good example of why it's important to be really careful when using familiar notation for unfamiliar things:

We're used to the symbols <=, >= being used to express relationships between real numbers, and when they are used in this context, it's true that if a<=b and b<=a, then a=b.

In the context of the OP, the symbols <=, >= are being used to express relationships between cardinal numbers (which are essentially the different sizes that sets can have), and the way that <= and >= are defined is different to their definition as relations on the set of real numbers:

If A and B are sets, we say |A|<=|B| if there is an injective function from B to A, and say |A|>=|B| if there's an injective function from A to B. We also say |A|=|B| if there's a bijective function from A to B.

If A and B are finite sets, then if |A|>=|B| and |B|<=|A|, it's easy to show that |A|=|B|. However, for infinite sets, this is a lot harder to show, (for example, the set of natural numbers N obviously injects into the set of rational numbers Q, and you can inject Q into N by mapping the fraction a/b in its lowest terms to 2^a 3^b, but it's not necessarily obvious that a bijective mapping exists) and that's what this question is about.

I am really sorry, I did mean it for finite sets yes. I meant o write it I believe, my bad. For infinite sets, we will have to set up a bijextion as you (or another user) commented

The statement is equivalent to Cantor-Bernstein:

Suppose the statement holds, and let the following maps be injective $\displaystyle M _g\leftrightarrows^f N$. Then g(M) is a subset of N, and gf is an injection from N to g(M), so by from the statement, there's a bijection $h: g(M) \rightarrow N$. So hg is bijection between M and N, so Cantor-Bernstein follows.

Basically, you're being asked to prove something equivalent to Cantor-Bernstein, so you're going to end up having to "copy" the proof somehow.