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I am so sorry for posting about it more than one time but this is my 4th revision for this proof and I want some feedback..

Note : $[n]$ is $\{1,2,3,4,\ldots,n\}$, a finite set

Prop: Every subset of a finite set is finite

Proof: Let $A,B$ sets and suppose $B$ is a finite set.

Also let $A \subseteq B$

If $A = \varnothing$, then it is finite

If $B = \varnothing$, then also $A = \varnothing$, which is finite.

So we suppose $A \neq \varnothing$ and $B \neq \varnothing$. Then there exstis a bijection $f: B \to [n]$

Since $A \subseteq B$, the inclusion mapping $I : A \to B$ defined on $f(a) = a$ for $a \in A$ exists.

Note $I(a_1) = a_1$ and $I(a_2)=a_2$ and if we suppose $I(a_1)=I(a_2)$, then $a_1 = a_2$, and therefore $I$ is an injection.

Then we can have $f \circ I: A \to [n]$, which is injective.

We need to prove that there is a bijection between the set $A$ and subset of $[n]$

So, let $p(n)$ be the statement that "If $A \subseteq [n]$, $A$ has a bijection with $[m]$ for some $m \in Z_{\ge0}$ such that $m \le n$"

Base case $(n=1)$ : If $A \subseteq [1]$, since we suppose $A \neq \varnothing$, $A$ has a bijection with $[1]$

Induction part : Now let $k \in N$ and suppose $p(k)$ holds. Now we want to prove that $p(k+1)$, that is, "If $A \subseteq [k+1]$, $A$ has a bijection with $[m]$ for some $m \in Z_{\ge0}$ such that $m \le k+1$"

Then if $A $ has $k+1 $ elements, it is done. It has a bijection with $[k+1]$.

If $A \subseteq [k]$, by the induction hypothesis, $A$ has a bijection with $[m]$ for $m \le k$.

Hence every subset of finite set is finite.

Michael Hardy
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Meer
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  • Related: http://math.stackexchange.com/questions/239566/subset-of-a-finite-set-is-finite – Asaf Karagila May 09 '16 at 21:58
  • so in your proof, there are two parts. Is it that I need to prove one of those two? or both two? It seems my proof is similar to your first proof, but I wonder if I also have to do the second one – Meer May 10 '16 at 00:42
  • @AsafKaragila do you see anything wrong with my proof above though for now? – Meer May 10 '16 at 00:56

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Would this be easier? An infinite set $S$ trivially has infinitely many subsets (take e.g. the singletons $\left\{ s \right\}$, $s \in S$). By a simple combinatorial argument a finite set $F$ has $2^{\#F}< \infty$ subsets.

Assume that $A \subset F$ is infinite, with $F$ a finite set. Then there are infinitely many $B \subset A \subset F$, which is a contradiction.

Matias Heikkilä
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  • Thanks for your comment but I was told that I have to use an induction to prove it.. but thanks!!! – Meer May 09 '16 at 21:24
  • If you need to use induction, then use induction on the size of the set, so start with a set of 1 element, then go on to a set An with n elements. If C is a subset of An, then C either contains the element 'an' , or it doesn't . If C does'nt contain the new element, then C is essentially a subset of of An-1, so we can define a bijection fC which maps C to [n-1] so by the induction hypothesis is finite. Otherwise C = B U {an} where B is a subset of An-1, so we can define a map fC which maps elements of B to the [n-1] and maps an to n, hence fC is a bijection from C to [n], so C is finite. – user247608 May 10 '16 at 02:50