Proving that the set of subsets of size $n$ of an infinite set $A$ has greater cardinality than (or equal to) the set $A$. I was going through some of the exercises of Lang's Algebra and stumbled upon the Set theory appendix. The question is the following : 

If $A$ is an infinite set, and $\Phi_n$ is the set of subsets of $A$ having exactly $n$ elements, show that $\mathrm{card}\,A\leq\mathrm{card}\,\Phi_n$. 

I thought of proving that statement by induction by defining a injective function from $\Phi_n$ to $\Phi_{n+1}$ but there's a problem with the mapping I came up with as I'll describe below. 
There is an easy bijection from $A$ to $\Phi_1$ that maps $x\in A$ to $\{x\}\in\Phi_n$ so the statement is true for $n=1$. 
Assume the statement is true for some $n\in\mathbb{N}$. Define the function $f_a:\Phi_n\to \Phi_n\cup\Phi_{n+1}$ such that $f_a(X)=X\cup\{a\}$ with $a\in A$. One can show that the function is indeed injective so we have that $\mathrm{card}\,\Phi_n\leq \mathrm{card}\,\Phi_n\cup\Phi_{n+1}$. At first I meant for the codomain of $f_a$ to be $\Phi_{n+1}$ which would have allowed me to finished the proof because then I could show that $\mathrm{card}\,A\leq\mathrm{card}\,\Phi_n\leq\mathrm{card}\,\Phi_{n+1}$ but that raises some issues. Indeed for $X\in\Phi_n$ such that $a\in X$ then $f_a(X)=X\notin \Phi_{n+1}$ which is why I defined the codomain as $\Phi_n\cup\Phi_{n+1}$. So is there a way to salvage my approach or am I going the wrong way ? 
 A: The approach can be modified, but to what extent this is a salvation or not is up to you.
Let me deal with the case of $n=2$, and you can think how it generalises further, and there are several "natural ways" to generalise, see if you can come up with more than one.
Fix $a$ and fix some $b,c\neq a$. Now define the following function:
$$F(x)=\begin{cases}\{a,x\} & a\neq x\\\{b,c\} & a=x\end{cases}$$

Another way of doing this is abstractly by means of cardinal arithmetic.
The problem you have here is on a finite set. Only the elements of $X$ are problematic. So this is an injection from $A\setminus X$ into $[A]^n$, which is the set theoretic notation of $n$-elements subsets. But as $A$ is infinite and $X$ is finite, it follows that $|A|=|A\setminus X|$.
Of course this is not only less satisfying, but also requires the result that $|A|=|A\setminus X|$, which depends to some extent on the axiom of choice, compared to the first way which is more hands-on, and only requires that $A$ has more than $2$ elements (or generally, more than $n^2$ elements or so).
