If $A$ is an infinite set, prove that $A$ has a proper infinite subset I considered taking this contradictory approach: Let $A_1\in A$. Assume that the proper subset $A\smallsetminus\{A_1\}$ is finite such that there exists a bijection $f:A\smallsetminus\{A_1\} \to C_k$ with $C_k=\{1,2,3,\ldots,k\}$ with $k \in N$.
What I wanted to do next is somehow prove that since $A\smallsetminus\{A_1\}$ is finite, then $A$ is finite, contradicting the given and proving my assumption false. However I have no idea how to proceed from here, or even if what I've done so far is plausible. Any help please?
 A: Extend the mapping to $C_{k+1}$ by mapping $A_1$ to $k+1$. Then the resulting mapping $A \mapsto C_{k+1}$ is a bijection, and as $C_{k+1}$ has cardinality $k+1$, it follows that $A$ must have cardinality $k+1$ a finite number as well.
A: $f$ is a bijection. Consider the function $g$ which is initially a copy of $f,$ but then we are also going to append the element $A_1$ to the domain and append $g(A_1)=k+1$ to the codomain.
Then, $g:A\to C_{k+1}\ $ is a bijection with $k+1\in\mathbb{N}$ elements in the codomain.
This shows that $A$ has finite cardinality, which gives you the contradiction you were after.
A: One can remove a finite set of elements from that infinite set without changing it's cardinality.
$S$ be an infinite set and $A  \subset S$ be any finite set.
Claim : $Card(S)=Card(S\setminus A)$
Proof: Since $A$ is finite, assume $$A\sim\mathbb{N_n}$$ for some $n\in \Bbb{N}$.
Let, $A=\{x_1,x_2,,...,x_n\}$
Then by Axiom of choice, there exists a choice function
$$f: {\scr{P}}(S)\to S $$
such that $f(E) \in E \space$,  $\emptyset \neq E\subset X$
Then define,
\begin{align}x_{n+k} &=f(S\setminus \{x_1,x_2,...,x_{k}\})&k=1,2,... \end{align}
Now, define a map $$h:S\to S\setminus A$$ by
\begin{equation}
  h(x) =
    \begin{cases}
      x_{n+k} & \text{; } x=x_k\\
      x & \text{ otherwise }
    \end{cases}       
\end{equation}
Then, $h$ define a bijection.
Hence, $Card(S) =Card(S\setminus A) $
Hence, $S$ has a proper infinite subset $S\setminus A$.
A: I must assume that you mean that there exists a subset $B\subsetneq A$ such that it is bijective with $A$. Because otherwise it would only be enough to remove a point or two and you would have what you want: let $x\in A$, then $B=A\setminus\{x\}\subsetneq A$. End.
Since $A$ is infinity, then there exists $f:\mathbb N\to A$ injective. We take $X=f(\mathbb N)\subset A$. then $X=\{a_1,a_2,\dots\}$ with $a_i\neq a_j$ if $i\neq j$. Let $B=(A\setminus X)\cup \{a_2,a_4,\dots,a_{2n},\dots\}\subset A$. Now let's define $g:A\to B$, given that $$f(a)=\left\{\begin{array}{ccl}a&,&a\in A\setminus X\\
a_{2n}&,&a=a_{n}\end{array}\right.$$
It is clearly surjective, let's see that $g$ is injective. We suppose that $g(y)=g(z)$.
If $y,z\in A\setminus X$, then $y=g(y)=g(z)=z$.
If $y,z\in X$, then $y=a_n$ and $z=a_m$ for some $n,m\in\mathbb N$. So $g(y)=g(z)\Rightarrow a_{2n}=a_{2m}\Rightarrow 2n=2m\Rightarrow n=m \Rightarrow a_n=a_m\Rightarrow y=z$.
If $y\in A\setminus X$ and $z\in X$, then $y=g(y)=g(z)=a_n$ for some $n$, so $y=a_n\in X$. Contradiction.
Hence $g$ is a biyection between $A$ and $B\subsetneq A$.
