If a set $A$ is finite then $A\cap B$ is a finite set.

Question

Background:

Theorem - If $A\subseteq \mathbb{N}_n$ then $A$ is a finite set and $|A|\leq n$.

Question:

Let $A$ be a finite set and $B$ be some set. If $A$ is a finite set, then $A\cap B$ is a finite set.

Attempted proof:

Let $A$ be a finite set and let $C = A\cap B$. If $C = \emptyset$ then $C$ is finite. Suppose $C \neq \emptyset$, since $C\subseteq A$, the set $A\neq \emptyset$ and there exists $k\in\mathbb{N}_k$ such that $A\sim \mathbb{N}_k$. That is, there exists $k\in\mathbb{N}$ and there exists a one-to-one correspondence $f:A\to\mathbb{N}_k$. The restriction of $f$ on the set $C$ $f|_C$ is a one-to-one function from $C$ onto $f(C)$. Therefore, $C\sim f(C)$. By theorem above, some $f(C)$ is a subset of $\mathbb{N}_k$, $f(C)$ is a finite set therefore since $C\sim f(C)$, $C$ is finite as well.

I am not sure if this is completely right, any feedback or other approaches would be appreciated.

Answer 1

Hint $:$ $A \cap B \subseteq A.$

Answer 2

Since $A\cap B\subseteq A$, we have an obvious injection

$$A\cap B \to A$$

Answer 3

1) $A$ is finite: Then $A\cap B$ is finite,

since $A \cap B \subset A$.

Let $A$={$x_1,x_2,.....,x_n$}$, x_i$ are distinct , $1 \le i \le n$, $n$ is the number of elements of $A$.

Let $n_1$ be the smallest positive integer s.t.

$x_{n_1} \in A\cap B$.

Let $n_2$ be the smallest pos.. integer $n_2 >n_1$ s.t.

$x_{n_2} \in A\cap B$.

Let $n_l$ be the smallest pos. integer $n_l >n_{l-1}$ s.t.

$x_{n_l} \in A\cap B$.

This process comes to an end after $m \le n$ steps.

$A\cap B=${$x_{n_1},x_{n_2},. x_{n_m}$}.

Left to do:

Find a bijection $A\cap B$ $ \rightarrow $ $J_m$, where $J_m=${$1,2,...m$}.

Answer 4

Suppose $A\cap B$ is infinite. Then we have $x_1,x_2,x_3,....$ all in $A\cap B$. But then each of them is also in $A$. Then $A$ has ininite cardinality. A contradicition.